Solutions to The Rising Sea (Vakil)

Solution:

1. Suppose \(\CC\) is a groupoid with one object \(\textrm{Ob}(\CC) = \{ A\} \). Then there is a group structure on \(\Mor_{\CC}(A, A) \), since the existence of \( \textrm{id}_A \in \Mor_{\CC}(A, A) \) is guaranteed by our axioms, and by definition of a groupoid every morphism is an isomorphism \(\Rightarrow\) \(\forall g \neq \textrm{id}_A \in \Mor_{\CC}(A, A), \exists g^{-1} \in \Mor_{\CC}(A,A) \).
2. Take a groupoid with \(\textrm{Ob}(\CC) = \{\Z, D_6\} \) and suppose $$ \Mor_{\CC}(\Z,\Z) = \textrm{Aut}(\Z) \\ \Mor_{\CC}(D_6,D_6) = \textrm{Aut}(D_6) \\ \Mor_{\CC}(\Z,D_6) = \emptyset $$ Both endomorphism rings have a group structure (clearly as automorphism groups), but the overall groupoid cannot be isomorphic to a groupoid with one object since there is no identity. This is the same as saying the disjoint union of groups is not a group.

Proof:

Proof:

Fix some finite dimensional vector spaces \(V, W \) over \(k\) (need not be the same dimension), and suppose \(A : V \to W\) is a linear transformation. We define our natural maps \(m_V : V \to V^{\vee \vee} \) and \( m_W : W \to W^{\vee \vee} \) by \(m_V(v) = (V^\vee \ni f \mapsto f(v) ) \) and \(m_W(w) = (W^\vee \ni g \mapsto g(w)) \). Recall that we may define the dual of our linear transformation \(A^\vee : W^\vee \to V^\vee \) by

for \(g \in W^\vee \). In a similar fashion, we may define \(A^{\vee \vee} : V^{\vee \vee} \to W^{\vee \vee}\) by

Suppose we fix some \(g \in W^\vee \). Then for any \(v \in V\) we have

In particular, for any \(v \in V\), both \( A^{\vee\vee} \circ m_V(v) \) and \(m_W(Av)\) are precisely the \(k\)-linear map \(g \mapsto g(Av)\) so that the following diagram commutes

In other words, the map \(m_* : \mathscr{Id} \to (\cdot)^{\vee\vee} \) is a natural transformation of functors. Recall from elementary linear algebra that, for a finite dimensional vector space \(V\), the transformations \( m_V : V \to V^{\vee \vee} \) are in fact isomorphisms (for infinite-dimensional vector spaces the map need only be injective, so that \(V\) embeds into \(V^{\vee\vee} \)).

The proof is fairly easy, but to see that \( m_V \) is always injective, suppose that \( v \in \ker m_V \). Then for all \( f \in V^\vee \), \(f(v) = 0\) (as \(m_V(v)\) is the zero map). By linearity, this implies \(v = 0\).

Now when we know \(V\) is finite dimensional, \(V^\vee\) must also have the same finite dimensions (this simply amounts to showing the dual of a basis in \(V\) is a basis in \(V^\vee\)). Repeating this argument, \( \dim(V) = \dim(V^{\vee\vee} \) — since an injective map between vector spaces of the same dimension is necessarily surjective, we have that \(m_V\) is an isomorphism. Therefore, the double dual functor is naturally isomorphic to the identity functor.

Solution:

From the preceding paragraph we have that \( \mathscr{V} \) is the category whose objects are the \(k\)-vector spaces \(k^n\) for \(n \geq 0\) (so one object for each \(n\)) and morphisms are linear maps \(k^n \to k^m \) — this category clearly a subcategory of \(f.d.\textrm{Vec}_k\), so that the functor \( \mathscr{V} \to f.d.\textrm{Vec}_k\) is just the inclusion functor \(\iota\), which is always faithful.

Recall from linear algebra that two vector spaces \(V\) and \(W\) of the same finite dimension are necessarily isomorphic — unfortunately the isomorphism here requires fixing a basis, so there is some loss of naturality. However, using the terminology of the following paragraph in Vakil, this shows that every finite-dimensional vector space \(V\) is isomorphic to some \(k^n\) (so that our functor is essentially surjective, as later defined). To see that the functor is full, let \(T : V \to W\) be a linear map between two finite-dimensional vector spaces \(V\) and \(W\), with \(n = \dim(V)\) and \(m = \dim(W) \). By fixing bases for \(V\) and \(W\) (and hence representing \(T\) as a matrix), there exist isomorphisms \(\phi : k^n \to V \), \(\psi : W \to k^m \) so that \(\iota(\psi \circ T \circ \phi) \) is a similar matrix to \(T\). Since similar matrices represent the same linear transformation, it should be clear that the functor is fully faithful.

This allows us to define our "inverse" functor (which Vakil mentions is really a quasiinverse)

Indeed, this map is only well-defined + unique up to isomorphism.

Proof:

Let \(A, A'\) be two initial objects of a category \( \CC \). Since \(A\) is initial, there exists a single (unique) map \(\phi : A \to A'\) — similarly, there is a single (unique) map \(\psi : A' \to A\), as \(A'\) is also initial. Now since \( \Mor_\CC(A, A) \) is a singleton and we always have the existence of an identity morphism, \( \Mor_\CC(A,A) = \{id_A \} \). By our composition law, we must have \( \psi \circ \phi \in \Mor_\CC(A,A) \), so it must be the case that \( \psi \circ \phi = id_A \) as that is the only map in \( \Mor_\CC(A,A) \). By an identical argument, \( \phi \circ \psi = id_{A'} \) since \( \Mor_\CC(A', A') \) only consists of the identity. Thus, \(A'\) and \(A\) are isomorphic.

The argument remains the exact same for final objects, with the only change being that the directions of our arrows are reversed.

Solution:

In \( \textbf{Set} \), we have that for any set \(A\), a function \( \varphi : \emptyset \to A \) can be interpreted as a subset of \( \emptyset \times A = \emptyset\) satisfying \(\forall x \in \emptyset \), \(\exists !y \in A\ \textrm{with}\ (x, y) \in \emptyset \times A \). Since the condition is vacuous and \( \emptyset \) is clearly the only subset of \( \emptyset\), this map is unique so \( \emptyset\) is initial.

However, if we flip the domain and codomain so that we are looking at maps \( f: A \to \emptyset \), the condition no longer holds vacuously, as every \( a \in A \) does not have an image. Therefore, the final object of \( \textrm{Set}\) is the singleton \( \{\emptyset \} \) since every point from our domain must map to it, giving the only possible map.

If we call our singleton point from the preceding paragraph \( 0 \), it follows that for every ring \(R\), there is a unique set-theoretic map \(f\) from the underlying set \(R\) to \( \{0\} \). By endowing \( \{0\} \) with the zero-ring structure so that \(0\) is both the additive and multiplicative identity, \(f\) trivially satisfies the definition of a ring homomorphism so that the zero ring is final in \( \textbf{Ring} \). However, the zero ring cannot be initial in \( \textbf{Ring} \) since for any ring \(S\) and ring homomorphism \( f : \{ 0 \} \to S \), \(f\) must preserve the additive and multiplicative identities \( \Rightarrow f(0) \) must be the additive and multiplicative identity \( \Rightarrow S\) is also the zero ring. Instead, recall that for any ring \(R\) there is a unique ring map \(\phi : \Z \to R \) given by

(This ring map is in fact how one defines the characteristic of a field). The uniqueness of this map shows that \( \Z\) is initial in \( \textbf{Ring} \).

In the category \( \textbf{Top} \), our initial and final objects are the same as \( \textbf{Set} \) for the following reasons: (1) Every topological space has an underlying set, so our set-theoretic maps exist; (2) The empty set \(\emptyset\) is always open, so any map \( \phi:\emptyset \to X \) is trivially continuous; (3) the preimage of our singleton map is the whole space \(X\) which is always open.

In the first category mentioned in Example 1.2.9, where we are looking at subsets of \(X\) such that there is a single morphism from \(U \) to \(V\) iff \(U \subset V\), it should be obvious that the initial object is \( \emptyset \) and the final object is our entire space \(X\). The same holds for our second category of open subsets of a topological space \(X\), since for any open set \(U \subset X\), there is exactly one inclusion \( \iota: U \to X \). The inclusion is clearly continuous since if \(Y \subset X \) is open, then \( \iota^{-1} (Y) = Y \cap U \) which is open by our topological axioms.

Proof:

Suppose \(S\) contains some zerodivisor \(a\), such that there exists a non-zero \(b \in A \) with \(ab = 0\). We wish to show \( \iota(b) = \frac{b}{1} \sim \frac{0}{1} = \iota(0) \). By construction of our equivalence relation \(\sim\) on \(A \times S\), we have

Thus \( b \neq 0 \) is in the kernel of our inclusion \( \iota : A \to S^{-1}A \) so that \(\iota\) is not injective.

Next suppose \(S\) contains no zero divisors and \( \iota(a) = \iota(b) \). Then \( \frac{a}{1} \sim \frac{b}{1}\), so there exists some \(s \in S\) with

Since \(S\) does not contain zerodivisors by assumption, it must be the case that \( a - b = 0\) or \(a = b \Rightarrow \iota \) injective.

Proof:

Suppose \(B\) is an \(A\)-algbra such that every element in \(S \subset A\) is mapped to an invertible element in \(B\) — let \( \phi : A \to B \) denote our ring map. Define the ring map \( \psi : S^{-1}A \to B \) by \( \psi(\frac{a}{b}) = \phi(a)\phi(b)^{-1} \) so that the following diagram commutes

We must show \(\psi\) is well-defined. Suppose \( \frac{a}{b} \sim \frac{c}{d} \) such that there exists some \(s \in S \) with

Since \( \phi : A \to B \) is a ring map, this gives us

Since \( \phi(s) \) is invertible by assumption, we have \( \phi(a)\phi(d) = \phi(b)\phi(c) \Rightarrow \psi(\frac{a}{b}) = \psi(\frac{c}{d})\). If we assume \(B\) is a commutative ring (which is the case in Vakil), the map \(\psi\) is indeed linear and thus the unique ring morphism required.

Proof:

The hint clearly gives the existence of our set \( S^{-1}M \), so we need to check that it satisfies the universal property. Note that the inclusion map \(\iota: M \to S^{-1}M \) is \(m \mapsto m/1 \) — we wish to show that if \(\phi : M \to N \) is another \(A\)-module map where \(N\) is also an \(S^{-1}A\)-module, \(\phi\) factors through \( \iota \). We will assume without loss of generality that \(S\) has no zero divisors, as otherwise our module \(S^{1}M =\{0 \}\) in which case the statement trivially holds.

Suppose \(\phi : M \to N\) is \(A\)-linear, and our \(A\)-mod structure on \(N\) given by \(\rho :A \times N \to N \) can be extended to an \(S^{-1}A\)-mod structure \(\widetilde{\rho} : S^{-1}A \times N \to N \). Define a map \( \psi : S^{-1}M \to N \) by

By module axioms, \( \psi \circ \iota (m) = \psi(\frac{m}{1}) = \rho(\frac{1}{1}, \phi(m)) = \phi(m) \), so if we can show this map is actually well-defined then it uniquely defines the ring map required for our universal property. In this light, suppose \( \frac{m_1}{s_1} = \frac{m_2}{s_2} \) in \(S^{-1}M\). By explicit construction in the hint, this implies that there exists some \(s \in S \) with

Since \(\phi\) is at the very least \(A\)-linear, this implies that \( \phi( s \cdot (m_1 s_2 - m_2 s_1) ) = 0 \) or equivalently

Since \(S\) is assumed to have no zerodivisors, the map \(\psi\) is well-defined.

Proof:

1. Applying induction, we need only check \( S^{-1}(M_1 \oplus M_2) \cong S^{-1}M_1 \oplus S^{-1}M_2) \). Define \( \phi : S^{-1} (M_1 \oplus M_2) \to S^{-1}M_1 \oplus S^{-1}M_2 \) by

   $$ \dfrac{(a, b)}{s} \mapsto \left( \frac{a}{s}, \frac{b}{s} \right) $$

   Notice

   $$ \begin{align} \phi\left( \dfrac{(a, b)}{s} + \dfrac{(c,d)}{t} \right) &= \phi\left( \dfrac{t(a, b) + s(c, d)}{st}\right) \\&= \left( \dfrac{ta + sc}{st}, \dfrac{tb + sd}{st} \right) \\&= \left( \dfrac{a}{s}, \dfrac{b}{s} \right) + \left( \dfrac{c}{t}, \dfrac{d}{t} \right) \\&= \phi\left( \dfrac{(a,b)}{s} \right) + \phi\left( \dfrac{(c,d)}{t} \right) \end{align} $$

   Additionally, it is easy to show that \( \phi (r \cdot \dfrac{(a, b)}{s} ) = r \cdot \phi(\dfrac{(a, b)}{s}) \) for any \(r \in A\) (this linearity may in fact be extended to \(S^{-1}A\)).

   To see that \( \phi \) is surjective, fix some \( \left( \frac{a}{s}, \frac{b}{t} \right) \in S^{-1}M_1 \oplus S^{-1}M_2 \). Then \(st \in S\) since \(S\) is a multiplicatively closed set, and thus

   $$ \phi\left( \dfrac{(ta, sb)}{st} \right) = \left( \dfrac{a}{s}, \dfrac{b}{t} \right) $$

   Lastly, we wish to show that \( \phi \) is injective. Suppose \( \phi\left( \dfrac{(a,b)}{s} \right) = \left(\dfrac{a}{s}, \dfrac{b}{s} \right) = \left( \dfrac{0}{1}, \dfrac{0}{1} \right) \). Then there exists some \(t_1 \in S\) with \(t_1 \cdot (a \cdot 1 - 0\cdot s) = t_1 a = 0\), and also some \(t_2 \in S\) with \(t_2 b = 0\). Taking \(t = t_1t_2\), we have

   $$ t \cdot ( 1\cdot (a, b) - s\cdot (0, 0) ) = (t_1t_2 a, t_1t_2 b) = (0, 0) $$

   so that \( \dfrac{(a,b)}{s} \sim \dfrac{(0, 0)}{1} \).

$$\tag*{$\blacksquare$}$$

2. Similar to before, we define a map \( \phi : S^{-1} \bigoplus_i M_i \to \bigoplus_i S^{-1}M_i \) by \( \dfrac{(m_i)}{s} \mapsto \left( \dfrac{m_i}{s}\right)\). By the same argument as before, it is easy to check that this is \(S^{-1}A\)-linear.

   Fix \( \left( \dfrac{a_i}{s} \right) \in \bigoplus_i S^{-1}M_i \) — it is important to remember here that only finitely many of the terms may be non-zero. Thus, there exists some \(N \geq 1\) with \( \frac{a_i}{s_i} = 0 \) for \(i > N\). Take \(s = \prod_{i=1}^N s_i\) and let \( t_k = \prod_{i \neq k}^n s_i \) — clearly \(t_k = s\) for \(k > N\). Again, since \(S\) is a multiplicatively closed set, it is closed under finite products so that \(s \in S\) and \(t_k \in S\) for all \(1 \leq k \leq N\) as desired. Then

   $$ \phi \left( \dfrac{(t_i a_i)}{s} \right) = \left( \dfrac{t_i a_i}{s} \right) = \left( \dfrac{a_i}{s_i} \right) $$

   To see that \(\phi\) is injective, suppose \( \phi \left( \dfrac{a_i}{s} \right) = \left(\dfrac{a_i}{s}\right) = 0 \). Then for each \(i\), we have that \( \frac{a_i}{s} \cong \frac{0}{1} \), so there exists some \( u_i \in S\) with \(u_i a_i = 0\). Moreover, since \( (a_i)\) is an element of the direct sum, all but finitely many may be non-zero. As before, this implies there exists some \(N \geq 1\) with \(a_i = 0\) for \(i > N\). By taking \( u = \prod_{i=1}^N u_i \), notice

   $$ u \cdot (a_i) = \left( \prod_{j \neq i}^N u_j \cdot u_i a_i \right) = 0 $$ so that \( \dfrac{(a_i)}{s} \cong \dfrac{(0)}{1} \). $$\tag*{$\blacksquare$}$$
3. Let \(S = \Z - \{0\} \), and consider the canonical map \( \iota : S^{-1} \prod_{i=1}^\infty \Z \to \prod_{i=1}^\infty \Q \), given as usual by $$ \dfrac{(m_i)}{s} \mapsto \left(\dfrac{m_i}{s}\right) $$ To see that \(\iota\) cannot be surjective, consider the element \( (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots) \in \prod_{i=1}^\infty \Q \). Suppose to the contrary there exists some \(s \in S\) and sequence \( (k_n) \) with \( \iota(\dfrac{(k_n)}{s} = \left(\dfrac{k_n}{s}\right) = \left(\dfrac{1}{n} \right) \). Then \( n \cdot k_n = s \) for all \(n \geq 1\), or equivalently \(n \mid s\) for all \(n \geq 1\) which is impossible. $$\tag*{$\blacksquare$}$$

Solution:

We are able to give (but not fully prove) a slightly stronger statement: \( \Z/(m) \otimes_\Z \Z/(n) \cong \Z / (\gcd(m,n)) \). To show this, define a \(\Z\)-linear map \(\phi : \Z \to \Z/(m) \otimes \Z/(n) \) by

This map is clearly surjective since for any \(a \otimes b\), bilinearity gives us \( \phi(ab) = ab(1 \otimes 1) = a \otimes b \). If we are able to show that \( \ker \phi = (\gcd(m,n)) \), then the result will follow from the first isomorphism theorem. On that note, suppose \(x \in (\gcd(m,n))\) so that we can write \(x = am + bn\) for some \(a, b \in \Z\). Then

Conversely, the statement \( \ker \phi \subset (\gcd(m,n) ) \) is equivalent to saying that if \( \gcd(m,n) \not\mid y \), then \(y \notin \ker \phi \). However, we may refine this statement a bit by noting that if \( \gcd(m, n) \not \mid y \), we may write \(y = c\cdot \gcd(m,n) + r \) for some \( 0 < r < \gcd(m,n) \) — by the previous paragraph and linearity of \(\phi\), we have \(\phi(y) = \phi(r) \). Therefore, if we can show \( \phi(r) \neq 0 \) for all \( 0 < r < \gcd(m,n) \), we will have shown \( \ker\phi \subset (\gcd(m,n)) \).

For general \(m,n \in \Z\), a proof of this last statement would require the universal property of the tensor product — which has purposefully not been introduced yet. Luckily, for the example at hand we only need to compute one value of \( 0 < r < \gcd(10, 12) = 2\). That is,

Proof:

Suppose we label the first map \( g : M^\prime \to M \). It is easy to see that since \(f : M \to M^{\prime\prime} \) surjective, the map \( f\otimes id_N : M \otimes_A N \to M^{\prime\prime} \otimes_A N \) is surjective: fix \(b \otimes_A n \in M^{\prime \prime} \otimes_A N \) and let \(a \in M\) be such that \(f(a) = b\). Then \(f \otimes id_N (a \otimes_A n) = b \otimes_A n\).

The crux of our proof will be showing that \( \im (g \otimes id_N) = \ker(f \otimes id_N) \); forward inclusion will be straightforward, but reverse inclusion will be quite unintuitive, and rely on the universal property of both the quotient and the tensor product. While using the universal property of the tensor product is not so out of touch here (considering it is introduced immediately after this exercise), the universal property of the quotient (factoring through the kernel of the cokernel) requires some additional facts we have not seen in Vakil up to this point. Fortunately, \(A\)-mod will be our prototypical abelian category so we know that the quotient exists, but it is worth mentioning the author likely intended this problem to be tackled from a different angle.

To prove that \( \im (g \otimes id_N) \subseteq \ker (f \otimes id_N) \), we need only show \( (f \otimes id_N) \circ (g \otimes id_N) = 0 \). But $$ (f \otimes id_N) \circ (g \otimes id_N) = (f \circ g) \otimes (id_N \circ id_N) = 0 \otimes id_N = 0 $$ since our original sequence was exact.

To show \( \ker (f \otimes id_N) \subseteq \im (g \otimes id_N) \), we will outline the proof given in Algebra (Hungerford, p.210). Consider the natural projection map \( \pi : M \otimes N \to M \otimes N / \im (g \otimes id_N) \). Then by the universal property of the quotient, since \(f \otimes id_N \) has domain \( M \otimes N \), there is a unique morphism \( \alpha : B \otimes N / \im (g \otimes id_N) \to M^{\prime \prime} \otimes N \) with

If we can show \(\alpha\) is an isomorphism, then it will be immediate that \( \ker(f \otimes id_N) \subseteq \im (g \otimes id_N) \). To do this, we directly construct an inverse \(\beta : M^{\prime \prime} \otimes N \to M \otimes N / \im (g \otimes id_N)\). We will use the universal property of the tensor product here; define \( \widetilde{\beta} : M^{\prime \prime} \times N \to M \otimes N / \im (g \otimes id_N) \) by

where \(b \in f^{-1}(c) \). This is clearly \(A\)-bilinear by linearity of the quotient map, so we wish to check that \(\widetilde{\beta}\) is independent of choice of \( b \in f^{-1}(c) \).

Suppose \( b' \in f^{-1}(c) \), so by \(A\)-linearity we have \(f(b - b') = 0 \Rightarrow b - b' \in \ker f = \im (g)\). Tensoring with \(n\), this gives us

so that \( \widetilde{\beta} \) is well-defined.

Going back to the universal property of the tensor product, this tells us that there exists a unique morphism \(\beta : M^{\prime\prime} \otimes N \to M \otimes N / \im (g \otimes id_N) \) with

for any \(b \in M \) with \(f(b) = c\). To see that this is indeed an inverse for \(\alpha\), notice

and

Proof:

Suppose we have another pair \( (T', t') \) (with \(t' : M \times N \to T'\) an \(A\)-bilinear map) satisfying the universal property of the tensor product. Then since our original map \(t : M \times N \to T \) is \(A\)-bilinear, there is a unique \(A\)-linear map \(f : T' \to T\) with \( t = f \circ t' \). By the same argument, we can find an \(A\)-linear map \(g: T \to T'\) with \(t' = g \circ t\). Then

Thus, if we can show that \(t\) and \(t'\) are epimorphisms then, it will follow that \(f \circ g = id_T \) and \( g \circ f = id_{T^\prime} \). Now in the category of \(A\)-modules, we have that a map is an epimorphism if and only if it is surjective. But this is obvious since the tensor product is explicitly defined as a quotient of \(M \times N\) by a certain equivalence relation, so both \(t\) and \(t'\) may be thought of as quotient maps.

Since both \(f\) and \(g\) are unique, the isomorphism described is unique, so any \(A\)-module satisfying the universal property must be isomorphic to \(M \otimes N\) — thus the tensor product is unique up to a unique isomorphism.

Proof:

Suppose \(t :M \times N \to T \) is a bilinear map. Let \( \pi : M \times N \to M\otimes_A N \hspace{1em}\left( = M \times N / \sim \right) \) denote the natural projection. To see that \(t\) factors through \( \pi \),

define the \(A\)-linear map \( \psi : M \otimes_A N \to T \) by \( \psi(m \otimes n) = t(m, n) \). Checking that \(\psi\) is well-defined amounts to checking that \(t\) is invariant under \(\sim\); that is, it preserves equivalence classes. But this is obvious since

as desired.

Solution:

1. This is simply the extension of scalars functor (which will be later expressed as the adjoint of the restriction of scalars). To give \(B \otimes_A M\) the structure of a \(B\)-module, define

   $$ b \cdot (b' \otimes m) := (bb') \otimes M $$

   It is straightforward to check that this satisfies the standard module axioms. To see that it gives a functor, we wish to show that it (covariantly) transforms our morphisms. In particular, if \(M, N\) are \(A\)-modules and \( f : M \to N \) is an \(A\)-linear map, we take

   $$ (B \otimes_A — ) (f) := id_B \otimes f $$

   This is clearly \(B\)-linear since

   $$ (id_B \otimes f)(b \cdot (b' \otimes m)) = id_B (bb') \otimes f(m) = b \cdot id_B(b') \otimes f(m) = b(id_B \otimes f)(b' \otimes m) $$

2. This is much easier to prove using isomorphisms of the tensor product instead of an explicit construction for arbitrary elements. Recall that \(A \cong A \otimes_A A \) via \(a \mapsto a(1 \otimes 1)\). Similarly, it is easy to show (using the universal property of tensor product) that \(C \otimes_A D \cong D \otimes_A C\) since our ring \(A\) is assumed to be commutative.

   Let \(u_B : A \to B \) and \(u_C : A \to C \) be \(A\)-algebra morphisms. There is a natural \(A\)-module structure on both \(B\) and \(C\) given by \( a \cdot b := u_B(a)b \) and \( a \cdot c := u_C(a)c \). Since \(B\) and \(C\) are in fact (assumed to be commutative so that \(u_B(A) \subset Z(B) \) ) rings, they are in fact \(A\)-algebras. This gives us two \(A\)-linear maps \(m_B : B \otimes_A B \to B \) and \(m_C : C \otimes_A C \to C \) representing our multiplication maps. Then we equip \(B \otimes_A C\) with an \(A\)-algebra structure via:

   $$ \require{amscd} \begin{CD} A \cong A \otimes_A A @>{u_B \otimes u_C}>> B \otimes_A C \end{CD} \begin{CD} (B \otimes_A C) \otimes_A (B \otimes_A C) \cong (B \otimes_A B) \otimes_A (C \otimes_A C) @>{m_B \otimes m_C}>> B \otimes_A C \end{CD} $$

Solution:

The two maps are simply

Proof:

It suffices to show that \(M \otimes_A (\oplus_{i \in I} N_i)\) satisfies the universal property of the coproduct over \( M \otimes_A N_i \). That is, we wish to describe morphisms \( f_i : M \otimes_A N_i \to M \otimes_A (\oplus_{i \in I} N_i) \) such that for any collection of morphisms \( g_i : M \otimes_A N_i \to N \) there exists a unique \( g : M \otimes (\oplus_{i \in I} N_i) \to N \) with \( g \circ f_i = g_i \).

For each \( j \in I\), define a map \( \hat{f}_j : M \times N_j \to M \otimes_A (\oplus_{i \in I} N_i) \) by

so that \( \widetilde{n}_i = \delta_{ij} \,n \). Clearly \(\hat{f}_j\) is \(A\)-bilinear, so there exists a unique \(A\)-linear map \( f_j : M \otimes_A N_j \to M \otimes (\oplus_{i\in I} N_i ) \) with \( f_j( m \otimes n ) = \hat{f}_j(m,n) \) well-defined. Now let \(N\) be another \(A\)-module with \(A\)-linear maps \( g_j : M \otimes_A N_j \to N \) for each \(j \in I\).

Define \( g : M \otimes_A (\oplus_{i \in I} N_i) \to N \) by

Since only finitely many of the \(\{n_i \}_{i \in I}\) are non-zero, only finitely many of the \( \{m \otimes n_i\}_{i \in I} \) are non-zero, and thus (by \(A\)-linearity) the sum over our \( g_i(m \otimes n_i) \) is in fact a finite sum. Moreover, by our construction of \(f_j\), we have

As desired. Therefore, \( M \otimes_A (\oplus_{i \in I} N_i ) \) satisfies the universal property of the coproduct over \( M \otimes N_i \) so that there is a unique isomorphism of \(A\)-modules \( \psi : M \otimes_A (\oplus_{i \in I} N_i ) \to \oplus_{i \in I} M \otimes_A N_i \).

Proof:

Let \(X, Y, Z\) be sets, \( \alpha : X \to Z \), \( \beta : Y \to Z\) be functions. We wish to show that the set

satisfies the universal property of the fibred product. To start, define the maps \(\pi_X : W \to X \) and \( \pi_Y : W \to Y \) to simply be the restriction of the natural projection maps on \(X \times Y\) to \(W\) — that is, \( \pi_X(x, y) = x \) and \( \pi_Y(x, y) = y \). Then by construction of \(W\) and our maps \(pi_X, \pi_Y\), it should be obvious that

commutes. Let \(W'\) be another set with maps \( p_X : W' \to X \), \(p_Y : W' \to Y\) such that

commutes. Define the map \( \psi : W' \to W \) by \( \psi(w) = (p_X(w), p_Y(w) ) \). Then clearly \( p_X(w) = \pi_X \circ \psi (w) \) and \( p_Y(w) = \pi_Y \circ \psi(w) \). Therefore, \(W\) satisfies the universal property of the fibred product.

Proof:

Recall that our morphisms of \( \mathcal{Top}_X \) are simply inclusion maps \( i : U \hookrightarrow X \) where \(U \subset X\) is an open set. Then in order for us to have a commuting square

of morphisms, we must somehow have that \( U \times_X V \) is an open subset of both \( U\) and \(V\) alike. In fact, since all our maps are inclusions, we should have that inclusion into \(U\) gives the same point as inclusion into \(V\) --- thus, any point in \( U \times_X V \) should lie in both \(U\) and \(V\). It should be clear that \( U \times_X V = U \cap V \) satisfies the universal property, so that in \( \mathcal{Top}_X \) fibred products are really \( \textbf{intersections} \).

Proof:

We wish to show that the product \(X \times Y\) satisfies the universal property of the fibred product over \(Z\). By definition, the product \(X \times Y\) comes equipped with projection maps \( \pi_X : X \times Y \to X \) and \( \pi_Y : X \times Y \to Y\), such that any other object \(P\) with maps \( p_X : P \to X \) and \( p_Y : P \to Y \) must factor uniquely through \( X \times Y \).

Using the fact that \(Z\) is final, \( \Mor(X, Z) \) and \( \Mor(Y, Z) \) are singleton sets — there exist unique morphisms \( \phi_X : X \to Z \) and \( \phi_Y : Y \to Z \). Similarly, \( \Mor(X \times Y, Z) \) is a singleton set, so \( \phi_X \circ \pi_X, \phi_Y \circ \pi_Y : X \times Y \to Z \) must be the same morphism (as there is only one possibility). Therefore, the diagram

must necessarily commute. Now let \(W\) be another object with maps \( p_X : W \to X \) and \( p_Y : W \to Y \) such that

commutes. Ignoring the bottom-right corner of the diagram, we may utilize the universal property of the (usual) product to see that there exists a unique morphism \( \psi : W \to X \times Y \) such that \( \pi_X \circ \psi(w) = p_X(w) \) and \( pi_Y \circ \psi(w) = p_Y(w) \) — but this is the same required property for the universal property of the fibred product, so that the product also satisfies the universal property of the fibred product.

Proof:

Label the morphisms in our diagram as follows:

We first need to show that the outer diagram commutes before moving on to universal property nonsense. Since the inner two Cartesian diagrams commute, we get

Then \( (\alpha \circ f') \circ p_W = (\beta \circ f) \circ p_W = \beta \circ (f \circ p_W) = \beta \circ (g \circ p_V)\) as desired.

Next, let \(S\) be another object along with maps \( \pi_Y : S \to Y \) and \( \pi_V : S \to V \) so that the outside of our square commuties — that is, \( \alpha \circ \pi_Y = (\beta \circ g) \circ \pi_V \). Define \( \pi_X = g \circ \pi_V \in \Mor(S,X) \), so that we may rewrite our commutivity condition on \(S\) as \( \alpha \circ \pi_Y = \beta \circ \pi_X \). By using our universal property of the fibred product on the lower Cartesian diagram, there exists a unique morphism \( \psi_1 : S \to W \) such that \( \pi_Y = f' \circ \psi_1 \) and \( \pi_X = f \circ \psi_1 \). By expanding our definition of \( \pi_X \), we have made a new commuting diagram, since our last equality tells us \( g \circ \pi_V = \pi_X = f \circ \psi_1 \):

Using our universal property of the fibred product on the top Cartesian square, we get another unique morphism \( \psi_2 : S \to U \) such that \( \psi_1 = p_W \circ \psi_2 \) and \( \pi_V = p_V \circ \psi_2 \). Then unwinding all of our compositions, we get

In other words, \( \psi_2 \) uniquely factors through \( f' \circ p_W \in \Mor(U, Y) \) and \( f \circ p_V \in \Mor(U, V) \) as desired.

Proof:

Let us denote our maps \( f: Y \to Z \), \( \alpha : X_1 \to Y \) and \(\beta : X_2 \to Y \). Assume \( X_1 \times_Z X_2 \) exists so that there exist maps \( \pi_1 : X_1 \times_Z X_2 \to X_1 \) and \( \pi_2 : X_1 \times_Z X_2 \to X_2 \) with \( (f \circ \alpha) \circ \pi_1 = (f \circ \beta) \circ \pi_2 \)

Similarly, suppose \(X_1 \times_Y X_2\) exists, such that there exists maps \( p_1 : X_1 \times_Y X_2 \to X_1 \) and \( p_2 : X_1 \times_Y X_2 \to X_2 \) with \( \alpha \circ p_1 = \beta \circ p_2 \). But then \( (f \circ \alpha) \circ p_1 = (f \circ \beta) \circ p_2 \), so that by the universal property of our fibred product \(X_1 \times_Z X_2 \) there exists a unique morphism \( \psi : X_1 \times_Y X_2 \to X_1 \times_Z X_2 \) with

The diagram here is:

Proof:

Let us use the same labels for our morphisms as the previous exercise. Suppose \( Y \times_Z Y \) exists so that there exist maps \( r_1, r_2 : Y \times_Z Y \to Y \) with \( f \circ r_1 = f \circ r_2 \). Since \( \alpha \circ \pi_1 = \beta \circ \pi_2 \) in our original diagram, \( f \circ \alpha \circ \pi_1 = f \circ \beta \circ \pi_2 \); thus, the universal property of the fibred product \(Y \times_Z Y\), there exists a unique morphism \( \xi : X_1 \times_Z X_2 \to Y \times_Z Y \) with \( \alpha \circ \pi_1 = r_1 \circ \xi \) and \( \beta \circ \pi_2 = r_2 \circ \xi \) — notice that this implies \( r_1 \circ \xi = r_2 \circ \xi \).

By the previous exercise, we obtained a unique morphism \( \psi : X_1 \times_Y X_2 \to X_1 \times_Z X_2 \) such that \( p_1 = \pi_1 \circ \psi \), \( p_2 = \pi_2 \circ \psi \). Ultimately, this gives us a commuting diagram

Putting several pieces together, \( \alpha \circ p_1 = \beta \circ p_2 \), \( r_1 \circ \xi = r_2 \circ \xi \), \( p_1 = \pi_1 \circ \psi \), and \(p_2 = \beta \circ \pi_2 \), the paths along the outside are the same as the paths through the middle. More technically speaking,

The last step in defining our magic diagram is to define a right inverse for the \(r_i : Y \times_Z Y \to Y\) maps. This exists since the following diagram trivially commutes

Thus,

Since \( \alpha \circ p_1 = \beta \circ p_2 \) represent the same map, we may label this as \(q : X_1 \times_Y X_2 \to Y \), so that by the above string of equalities \( \Delta \circ q = \xi \circ \psi \), giving rise to our commuting diagram

If \(W\) is another object with morphisms \( s_1 : W \to Y \) and \( s_2 : W \to X_1 \times_Z X_2 \) (such that \( \Delta \circ s_1 = \xi \circ s_2\)), the uniqueness of our magic diagram follows from extending \(s_2\) to \(\xi \circ s_2\), so that we may instead use the universal property of \( X_1 \times_Y X_2 \) as the fibred product to show that there exists a unique \( \sigma : W \to X_1 \times_Y X_2 \) such that \( s_1 \) and \(s_2\) necessarily factor through \( X_1 \times_Y X_2 \) (showing one of the maps factors through will be immediate while the other will require diagram chasing).

Proof:

Let \(A\) and \(B\) be sets and define

Let \( i_0 : A \hookrightarrow A \coprod B \) be defined by \(i_0(a) = (a, 0)\) and \(i_1 : B \hookrightarrow A \coprod B \) similarly by \( i_1(b) = (b,1) \).

Suppose \(W\) is another set with maps \( j_0 : A \to W \) and \( j_1 : B \to W \). Then we may define our set-map \( \psi : A \coprod B \to W \) by \(\psi(x, c) = j_c(x)\). It is immediate from construction of our maps \(i_0, i_1\) that \( \psi \circ i_c = j_c \) for \(c = 0, 1\).

Proof:

Using the notation from exercise 1.3.K, let \( u_B : A \to B \) and \( u_C : A \to C \) denote the two \(A\)-algebra morphisms. As mentioned in the problem statement, we define \(A\)-linear maps \(\phi_B : B \to B\otimes_A C \) and \(\phi_C : C \to B \otimes_A C \) via \( b \mapsto b \otimes 1 \) and \(c \mapsto 1 \otimes c\), respectively. This diagram clearly commutes via the \(A\)-module structure on \(B\) and \(C\) since \( u_B(a) = a \cdot u_B(1_A) = a \cdot 1_B \) and \( u_C(a) = a \cdot u_C(1_A) = a \cdot 1_C \), giving us

Now suppose \(S\) is another \(A\)-algebra with \(A\)-linear ring morphisms \(\psi_B : B \to S \) and \( \psi_C : C \to C \) such that \( \psi_B \circ u_B = \psi_C \circ u_C \). Define the map \( \widetilde{f} : B \times C \to S \) by \( \widetilde{f}(b, c) = \psi_B(b)\psi_C(c) \). This is clearly \(A\)-bilinear, so there exists a unique \(A\)-linear ring morphism \( f : B \otimes_A C \to S \) with \( f(b \otimes c) = f(b, c) = \psi_B(b)\psi_C(c) \). Then clearly for any \(b \in B\),

since \( \psi_C \) is a ring map (and thus preserves multiplicative identity). A similar argument shows that \( \psi_C \) factors through \( B \otimes_A C \), so that \( B \otimes_A C \) satisfies the universal property of fibred coproduct.

Proof:

Let \( f : X \to Y \) and \( g : Y \to Z \) be monomorphisms, and suppose \(W\) is another object along with morphisms \( s_1, s_2 : W \to X \) such that \( (f \circ g) \circ s_1 = (f \circ g) \circ s_2 \). Since composition is an associative operator by our categorical axioms, we must also have \( f \circ (g \circ s_1) = f \circ (g \circ s_2) \). Since \(f\) is monic, we have \( g \circ s_1 = g \circ s_2\). Since \(g\) is monic, the result follows.

Proof:

For the forward direction, suppose \(\pi : X \to Y \) is a monomorphism, and consider the diagram

which trivially commutes. If \(W\) is another object along with morphisms \( \phi_1, \phi_2 : W \to X \) such that \( \pi \circ \phi_1 = \pi \circ \phi_2 \), then \(\pi \ \textrm{monomorphism} \Rightarrow \phi_1 = \phi_2 = \phi\) so that we may use \( \phi : W \to X \) to be the unique morphism "factoring" through all the necessary maps (this is obvious since the map we are factoring through is the identity.) Consequently, \(X\) satisfies the universal property of the fibred product, so up to a unique isomorphism we have \( X \cong X \times_Y X \).

Conversely, suppose \(X \times_Y X\) exists, with maps

and further assume that the induced map \( f: X \to X \times_Y X \) (induced from using the identity morphisms \(id : X \to X\) along the edges of the diagram) is an isomorphism. But then we must have \( \phi_1 = f^{-1} \) — since inverses are unique, this implies\( \phi_1 = \phi_2 \). To see that \( \pi : X \to Y\) is a monomorphism, let \( W \) be another object with maps \( \sigma_1, \sigma_2 : W \to X \) such that \( \pi \circ \sigma_1 = \pi \circ \sigma_2 \). Then by the universal property of the fibred product, there exists a unique morphism \( \psi : W \to X \times_Y X \) such that \( \sigma_1, \sigma_2 \) factor through \( X \times_Y X \) — that is \( \sigma_1 = \phi_1 \circ \psi \) and \( \sigma_2 = \phi_2 \circ \psi \). But since \( \phi_1 = \phi_2 = f^{-1} \), we necessarily have that \( \sigma_1 = f^{-1} \circ \psi = \sigma_2 \).

Proof:

Using the magic diagram from Exercise 1.3.S, since \( f : Y \to Z \) is a monomorphism, then by the previous exercise we have that \( Y \times_Z Y \cong Y \). Thus, we obtain a new Cartesian diagram:

By the universal property of the Cartesian diagram, there exists some unique morphism \( f^{-1} : X_1 \times_Z X_2 \to X_1 \times_Y X_2 \) such that \( f \circ f^{-1} = id \) (notice that this does not by itself prove that \(f\) is an isomorphism since we have not shown \(f^{-1}\) is a left-inverse; however, using the universal property will effectively drive this point home).

Proof:

1. Fix \( A \in \CC \) and let \( C \) range over \( \CC \). For any morphism \(f : B \to C \), there is an induced morphism of sets (in the case of locally small category)

   $$ f^* : \Mor(C, A) \to \Mor(B, A), \hspace{5em} g \mapsto g \circ f $$

   such that the following diagram commutes:

   $$ \require{amscd} \begin{CD} \Mor(C, A) @>i_C>> \Mor(C, A^\prime)\\ @VVf^*V @VVf^*V\\ \Mor(B, A) @>i_B>> \Mor(B, A^\prime) \end{CD} $$

   By interchanging the top row with \(C = A\) and the bottom row with \(B = C\), we will consider the image of the identity morphism \(id_A \in \Mor(A,A)\) (as it is always guaranteed to exist). For each \( u \in \Mor(C, A) \) we obtain a diagram of the same form as above:

   $$ \require{amscd} \begin{CD} \Mor(A, A) @>i_A>> \Mor(A, A^\prime)\\ @VVu^*V @VVu^*V\\ \Mor(C, A) @>i_C>> \Mor(C, A^\prime) \end{CD} $$

   To find our morpism \( g: A \to A' \), we simply take \( g := i_A(id_A) \). Then by commutivity of the diagram,

   $$ i_C(u) = i_C \circ u^* (id_A) = u^* \circ i_A (id_A) = u^* \circ g = g \circ u $$

   To see that the map \(g\) is necessarily unique, suppose \( h : A \to A' \) is another map satisfying \(i_C(u) = h \circ u\) for all \(C \in \CC\) and \( u \in \Mor(C,A) \). By taking \(C = A\) and looking at the map \( u = id_A \), we can see that by commutativity of the diagram

   $$ g = i_A(id_A) = h \circ id_a = h $$

   as desired.

2. Assume the \( i_C : \Mor(C, A) \to \Mor(C, A')\) are all isomorphisms. Then the inverse maps \( i_C^{-1} : \Mor(C, A') \to \Mor(C, A) \) satisfy the hypothesis of part (a) above, so there must exist some unique morphism \( \widetilde{g} : A' \to A \) such that, for all \( C \in \CC \), \(i^{-1}_C(\mu) = \widetilde{g} \circ \mu\). Taking \(C = A \) as we did above, we have a composition

   $$ \require{amscd} \begin{CD} \Mor(A, A) @>i_A>> \Mor(A, A^\prime) @>{i^{-1}_A}>> \textrm{Mor(A,A)} \end{CD} $$

   so that

   $$ id_A = i^{-1}_A ( i_A( id_A )) = i^{-1}_A(g \circ id_A) = \widetilde{g} \circ g \circ id_A = \widetilde{g} \circ g $$

   By an identical argument, we may substitute \(C = A'\) to get \( id_{A'} = g \circ \widetilde{g} \), so that \(g\) is indeed an isomorphism.

Proof:

1. Let \( j : h^A \Rightarrow h^B \) be a natural transformation, so that for every morphism \(f : C \to D \) we get a commuting diagram

   

   Taking \( f = id_A\) our diagram becomes

   $$ \require{amscd} \begin{CD} \Mor(A,A) @>j_A>> \Mor(B,A)\\ @VVid_*V @VVid_*V\\ \Mor(A,A) @>j_A>> \Mor(B,A) \end{CD} $$

   Giving us the morphism \( j_A(id_A) : B \to A \) as desired.

   Conversely, let \( \phi : B \to A \) be a morphism. For every object \( C \in \CC \), define

   $$ \tau(\phi)_C : \Mor(A, C) \to \Mor(B, C), \hspace{5em} \tau(\phi)_C(f) = h^B(f)(\phi) $$

   Then for any morphism \( g : C \to D \) and \( f \in \Mor(A, C) \), we get a commuting diagram

   $$ \require{amscd} \begin{CD} \Mor(A,C) @>\tau(\phi)_C>> \Mor(B,C)\\ @VVg_*V @VVg_*V\\ \Mor(A,D) @>\tau(\phi)_D>> \Mor(B,D) \end{CD} $$

   with

   $$ \begin{align} \tau(\phi)_D \circ h^A(f) &= \tau(\phi)_D(g \circ f) \\&= (g \circ f)_*(\phi) \\&= h^B(g) \circ h^B(f) (\phi) \\&= h^B(g) \circ \tau(\phi)_C (f) \end{align} $$

   In particular, this tells us that the above diagram commutes, and thus \( \tau(\phi) : h^A \Rightarrow h^B \) is a natural transformation.

2. This is very similar to part (a) above and will be put off until chapter 6 for the sake of not getting carpal tunnel.

3. Let us use \( \textrm{Nat}(h^A, \FF ) \) to denote the collection of natural transformations \( h^A \Rightarrow \FF \). Define a map \( \theta_{\FF,A} : \textrm{Nat}(h^A, \FF) \to \FF(A) \) by

   $$ \alpha \mapsto \alpha_A(id_A) $$

   We define our inverse as follows: fix \( a \in \FF(A) \). Then for every \(B \in \CC \), define the map

   $$ \tau(a)_B : h^A(B) \to \FF(B), \hspace{5em} \tau(a)_B(f) = \FF(f)(a) $$

   Then for any morphism \(g : B \to C \),

   $$ \begin{align} \FF(g) \circ \tau(a)_B(f) &= \FF(g) \left( \FF(f)(a) \right) \\&= \FF(g \circ f)(a) \\&= [\tau(a)_C \circ h^A(g)] (f) \end{align} $$

   so that we indeed get a commuting diagram

   $$ \require{amscd} \begin{CD} h^A(B) @>\tau(a)_B>> \FF(B)\\ @VVh^A(g)V @VV\FF(g)V\\ h^A(C) @>\tau(a)_C>> \FF(C) \end{CD} $$

   In other words, we have a map \( \FF(A) \to \textrm{Nat}(h^A, \FF) \) given by \( a \mapsto \tau(a) \). It remains to show that \(\tau\) and \( \theta_{\textrm{F},A} \) are indeed inverse.

   Fix \(a \in \FF(A) \). Then

   $$ \theta_{\FF, A} ( \tau(a) ) = \tau(a)_A (id_A) := \FF(1_A)(a) = id_{\FF(A) }(a) $$

   On the other hand, if \( \alpha : h^A \Rightarrow \FF \) is a natural transformation and \( f : A \to B \) is a morphism, then

   $$ \begin{align} \tau( \theta_{\FF, A}(\alpha) )_B(f) &= \tau(\alpha_A(id_A))_B (f) \\&= \FF(f) (\alpha_A(id_A)) \hspace{4em}(\textrm{by}\ \textrm{defn}) \\&= \alpha_B(h^A(f)(id_A)) \hspace{4em}(\textrm{by}\ \textrm{naturality}) \\&= \alpha_B( f\circ id_A) \\&= \alpha_B(f) \end{align} $$

Proof:

Recall a poset \((S, \geq)\) may be interpreted as a category \( \mathscr{I} \) whose objects are the elements of \(S\), and a single morphism \( x \to y \) iff \( x \geq y \). Since \( e \in \mathscr{I} \) is initial, for every \(x \in \mathscr{I}\) there exists a unique morphism \( e \to x \) (in other words, \( e \geq x \) for all \(x \in S\).)

Following the definition, suppose \( \CC \) is another category with functor \( \FF : \mathscr{I} \to \CC \); we will write \( A_i = \FF(i) \) for each \( i \in \mathscr{I} \). We claim \( \varprojlim A_i = \FF(e) \). Suppose \( \phi : j \to k \) is a morphism in \(\mathscr{I} \). Since \(e\) is initial, there exist unique morphisms \( f_j : e \to j \) and \( f_k : e \to k \). Moreover, it must be the case that \( \phi \circ f_j = f_k \) since \( \Mor(e,k) \) is a singleton. Since \( \FF(\phi \circ f_j) = \FF(\phi) \circ \FF(f_j) \), we get a commuting diagram

Proof:

Let \( \mathcal{L} \subseteq \prod_i A_i\) denote the set in the problem statement above. By definition / universal property of the product \( \prod_i A_i \), we have maps \( \pi_j : \prod_{i} A_i \to A_j \) for each \( j \in \mathscr{I} \) given by \( \pi_j ( \{ a_i \}_{i \in \mathscr{I}} ) = a_j \). Then the restriction of our maps \( \pi_j \) to \( \mathcal{L} \) satisfy the requirement that

commutes (by construction of \(\mathcal{L}\).) Moreover, \( \mathcal{L} \) must be universal with respect to this property since the product is indeed universal.

Solution:

1. Let the target category by \( \textbf{Ring} \) and let our index category\( \mathscr{I} \) be \( \mathbb{N} \), interpreted as a poset (such that there exists a single morphism \( f: a \to b \) iff \(a \mid b \) .) Letting \( \FF : \mathbb{N} \to \textbf{Ring} \) be the functor sending \( n \mapsto \frac{1}{n}\Z \), if \(f : a \to b\) is a morphism, then by taking \( r = b / a \in \mathbb{N} \) it should be clear that

   $$ \FF(f) : \frac{1}{a}\Z \hookrightarrow \frac{1}{b}\Z, \hspace{4em} [ \FF(f) ](m) = r \cdot m $$

   is the desired functor. Now for each \( k \in \mathbb{N} \), we get an inclusion \( i_k : \frac{1}{k}\Z \to \Q \). If \(Q'\) is any other ring with \(Z\)-linear maps \( g_k : \frac{1}{k}\Z \to Q' \) for each \( k \in \mathbb{N} \) (that commute with inclusions \( \frac{1}{a}\Z \hookrightarrow \frac{1}{b} \Z \) for \(a \mid b\) ), then we may simply define our morphism \( \psi : \Q \to Q' \) by \( \psi (p / q) = g_q(p) \). The fact that our \( g_k \) commute with inclusions ensures that \( \psi \) is well-defined with respect to the relation \( p/q \sim p'/q' \) iff \( pq' - p'q = 0 \) in \(\Q\).

2. Let \( \CC_X \) be the category \( \mathcal{P}(X) \) interpreted as a poset, with morphisms being the inclusion maps. Furthermore, suppose \( \mathscr{I} \) is an indexing category for our subsets \(A_i \subset X \). Then for each \( j \in \mathscr{I} \), there is an obvious inclusion \( f_j : A_j \hookrightarrow \bigcup_{i} A_i \). Moreover, if \( A_j \subset A_k \), the composition of inclusions is naturally an inclusion.

   If \( W \subset X \) is another subset with inclusion maps \( g_j : A_j \to W \) that commutes with \( A_j \subset A_k \) for all \( j, k \in \mathscr{I} \), then for each \(y \in \bigcup_i A_i \) there exists some \( r \in \mathscr{I} \) such that \(y \in A_r \). Thus, we may define a map \( \psi : \bigcup_i A_i \to W \) by \( \psi(y) = g_r(y) \). This is well-defined / independent of the choice of \(r\) such that \(y \in A_r\) since our \( g_i \) necessarily commute with inclusions \( A_j \subseteq A_k \).

Proof:

For ease of notation we will let \(X\) denote the set above, and for each \( j \in \mathscr{I} \) let \( \phi_k : A_k \to X \) be the natural inclusion \( a \mapsto ( a, k ) \). Additionally, let \( \FF : \mathscr{I} \to \textbf{Set} \) denote our functor (such that \( A_k := \FF(k) \) ). Fix some \( i, j \in \mathscr{I} \). Since \(\mathscr{I}\) is filtered, \(\exists k \in \mathscr{I}\) with maps \( \widetilde{f} : i \to k \), \( \widetilde{g} : j \to k \). Then there are induced maps \( f = \FF(\widetilde{f}) : A_i \to A_k \) and \(g = \FF(\widetilde{g}) : A_j \to A_k \) — in particular, \(X\) satisfies the universal property of the colimit iff the diagrams

commute, as specified in the construction of \(X \).

Now if \(W\) is any other set satisfying the colimit property, via some maps \( g_i : A_i \to W \), then we may define a set-theoretic maps \( \psi : X \to W \) by \( (a_s, s) \mapsto g_s(a_s) \). \( \psi \) is well-defined under \( \sim \) by construction of \(X\).

Proof:

Let \( \mathscr{I} \) be a filtered set and \( \FF : \mathscr{I} \to \CC \) be the functor describing the diagram. Moreover, let \( \hat{M} \) be the \(A\)-module

where again our addition is described as: for \( m_i \in M_i \), \( m_j \in M_j \), \( \textrm{filtered} \Rightarrow \exists l \in \mathscr{I}\), \(u : i \to l, v : j \to l\)

Our zero element is necessarily any \( (m_j) \in \hat{M} \) with \( u : i \to k \) such that \( \FF(u)(m_i) = 0 \) in \( M_k \) (i.e. eventually zero in the diagram). Notice that this does, in fact, give us the additive identity under the construction above, since if \( m_i \sim 0 \), \( \exists u : i \to k \) with \( \FF(u)(m_i) = 0 \). Now since \( \mathscr{I} \) is filtered, \( \exists l \in \mathscr{I} \) with arrows \( v_k : k \to l \), \( v_j : j \to l \). Then

Next, we wish to check that addition does not depend on our choice of target or arrows. Suppose, for \( i, j \in \mathscr{I} \), the filtration property yields \( l, l' \in \mathscr{I} \) with morphisms \( u : i \to l, v: j \to l \) and \( u' : i \to l', v' : j \to l' \). By the first filtration property, there exists \( k \in \mathscr{I} \) with \( w : l \to k, w' : l' \to k \) so that we get maps \(w' \circ v', w \circ v : j \to k \). By the second filtration property, there must be some other map \( \psi : k \to r \) such that

Letting \( f = \FF(\psi \circ w), g = \FF(\psi \circ w')\), it follows that \( f(y) = g(y') \) so that \( (y, l) \sim (y', l') \). A similar argument shows \( \FF(u)(m_i) \sim \FF(u')(m_i) \), so that addition is well-defined by construction of \(\sim\).

Scalar multiplication is much more straightforward since it only requires we set \( r \cdot (a_i, i)_{i \in \mathscr{I}} = (ra_i, i)_{i \in \mathscr{I}} \) for all \(r \in A\) — the \(A\)-linearity of our maps handles all well-definedness requirements. The fact \( \widetilde{M} \) is a colimit follows from the argument in the exercise above.

Solution:

The argument is effectively the same here, where one gives \(S\) the structure of a poset (in particular, there is a morphism \( s \to t \) if and only if \(\exists u \in A,\, t = us \)). Then we get a functor \( \FF : (S, \geq) \to \mathbf{A}-\textbf{Alg} \) given by \( s \mapsto A_s \), and using the notation for \(s \to t\) above, we get induced maps \( A_s \to A_t \) given by \( \frac{r}{s} \mapsto \frac{ur}{t} \). Note that our maps \( A_s \to S^{-1}A \) are simply the inclusion maps.

Solution:

For \( f \in \Mor(A, B) \), let \( g_* : \Mor(A, B) \to \textrm{A, B'} \) denote the map \( f \mapsto g \circ f \). Then we obtain the second half of the diagram:

Yielding a complete diagram

Proof:

Fix \( A \in \mathscr{A} \) and consider the identity morphism \( id_A : A \to A \). Then for any \(B \in \mathscr{B} \) and \( g \in \FF(A) \to B \), the following diagram commutes:

Using this, define \( \eta_A := \tau_{A, \FF(A)}( id_{\FF(A)} ) \). Then by commutivity of the diagram, \( g = g_*( id_{\FF(A)} ) = g \circ id_{ \FF(A) } \), so \( \tau_{A,B}(g) = \mathscr{G}g_*(\eta_A) = \mathscr{G}g \circ \eta_A \). Similarly, if \( f : A \to \mathscr{G}(B) \), we get a diagram

By analogously defining \( \epsilon_B := \tau_{\mathscr{G}(B),B }^{-1} (id_{\mathscr{G}(B)}) \), it follows by commutivity of the diagram that

as desired.

Proof:

Before proving the statement, it is worth noting that this (together with the next exercise) proves the tensor-Hom adjunction — this is probably the most widely used application of category theory, as it is effectively a more powerful version of currying. Alas, the proof will not rely on any of this real-world knowledge.

The easiest way to approach this problem is by constructing a map \( \Hom_A( M \otimes_A N, P ) \to \Hom_A(M, \Hom_A(N, P)) \) and showing it is bijective (as bijective morphisms are isomorphisms in the category \(A\)-mod), as opposed to giving an explicit construction of \( \Hom_A(M, \Hom_A(N,P)) \to \Hom_A( M \otimes_A N, P ) \). To start, let \( f : M \otimes_A N \to P \) be \(A\)-linear, and fix \( m \in M \). We define a map \( \tau(f)(m) : N \to P\) by

This is clearly \(A\)-linear since \(f\) is and the tensor product is linear in each component. In addition, by allowing \(m \in M\) to vary, the map \( m \mapsto \tau(f)(m)\) is also necessarily \(A\)-linear (by the same argument as before) and thus an element of \( \Hom_A(M, \Hom_A(N, P)) \). This gives us the desired map

Showing that \(f\) is injective is essentially showing that the data of two maps is the same no matter how we iterate over the arguments. To be more precise, if \( \tau(f) = \tau(g) \), then for all \(m \in M, n \in N\), \( \tau(f)(m)(n) = \tau(g)(m)(n) \Rightarrow f(m \otimes n) = g(m \otimes n) \) so that \(g = f\).

To see that \( \tau \) is surjective, fix \( \alpha \in \Hom_A(M, \Hom_A(N,P)) \) and define \( \phi_\alpha : M \times N \to P \) by

Then \( \phi_\alpha \) is clearly \(A\)-bilinear (as each of its arguments are \(A\)-linear) so by the universal property of the tensor product, there exists some unique \(A\)-linear map \(\widetilde{\phi_\alpha} : M \otimes_A N \to P \) such that \( \widetilde{\phi_\alpha}(m \otimes n) = \phi_\alpha(m ,n) \) for all \(m \in M, n \in N\). In particular, this gives us \( \tau ( \widetilde{\phi_\alpha} ) = \alpha \).

Proof:

If we let \(\FF\) denote the functor \( (\cdot) \otimes_A N \), then for any map \(f : M \to M' \) we define \( \FF(f) := f \otimes id_N \). Then our pullback map \( \FF(f)^* \) is given explicitly by

We wish to show

is a commuting diagram. To start, fix some \( \alpha \in \Hom_A(M^\prime \otimes_A N, P) \). For each \( m \in M\), \( \tau_M(\alpha)(m) \) is the map

However, this is the same as precomposing the map \( M^\prime \to \Hom_A(N, P) \) given by \( m \mapsto \tau_{M^\prime}(\alpha)(m) \) with \(f\), i.e.

so that the diagram commutes.

Proof:

It should immediately be evident that the tensor-hom adjunction is somehow in play, given that the extension of scalars is simply the functor \( \cdot \otimes_B A \) and the restriction of scalars requires a module homomorphism. Though the proof could in fact be accomplished using just this and a bit of knowledge about bimodules (to resolve the discrepancy between \( \Hom_B \) and \( \Hom_A \)), we will give a slightly more direct proof.

Let \( \phi \) denote the morphism of rings \(B \to A \), and let \( f : N \to M_B \) be a \(B\)-linear map. Define \( \widetilde{f} : N \times A \to M \) by \( (n, a) \mapsto a f(n) \); since \(f\) is \(B\)-linear and the second argument may be made \(B\)-linear (more specifically \(\phi^*A\)-linear), the universal property of the tensor product gives rise to a unique map \( \hat{f} : N \otimes_B A \to M \). But \( N \otimes_B A \) is clearly an \(A\)-module by \( a \cdot (n \otimes a') = n \otimes aa' \), so

so \( \hat{f} \) is \(A\)-linear, giving us a well defined map \( \tau : \Hom_B(N, M_B) \to \Hom_A(N \otimes_B A , M) \) defined by \( f \mapsto \hat{f} \). This is clearly injective since \( \hat{f} = \hat{g} \) implies \( af(n) = ag(n) \ \ \forall a \in A \Rightarrow f(n) = g(n)\) for all \(n\) — i.e. \(f = g\).

To see that \( \tau \) surjective, it is easier to use the tensor-hom adjunction from our previous exercise. That is, suppose \( \phi : N \otimes_B A \to M \) is an \(A\)-linear map. Then by the tensor-hom adjunction, there is a corresponding \( \psi : N \to \Hom_A(A, M) \). Now as \(B\)-modules, we have \( M_B \cong \Hom_A(A,M) \), which clearly inherits an \(A\)-module structure under \( \phi \).

Solution:

In an identical fashion to the restriction of scalars for ring maps \( B \to A \) from Exercise 1.5.E, every \( S^{-1}A \)-module can be considered as an \(A\)-module using the ring map \( a \mapsto \frac{a}{1} \), and every \( S^{-1}A\)-linear map \( f : S^{-1} N \to S^{-1} N'\) can be made an \(A\)-linear map via \( \iota(f)(a) = f(\frac{a}{1}) \) (we will use \( \iota : \Mod{S^{-1}A} \to \Mod{A}\) to denote the forgetful functor.) If we let \( S^{-1}( \cdot ) : \Mod{A} \to \Mod{S^{-1}A} \) denote the obvious localization functor \( M \mapsto S^{-1}M\), then any \(A\)-linear map \( g: M \to M' \) may be made \(S^{-1}A\)-linear via

It is important to notice here that the map \( S^{-1}g \) is well-defined, since if \( \frac{m_1}{s_1} \sim \frac{m_2}{s_2} \) there exists some \( b \in S \) with \( b (m_1 s_2 - m_2 s_1 ) = 0 \). By \(A\)-linearity of \(g\),

so that \( \frac{g(m_1)}{s_1} \sim \frac{g(m_2)}{s_2} \), as desired.

This association is precisely the bijection \( \tau_{M, S^{-1}N} : \Hom_{S^{-1}A}( S^{-1}[M], S^{-1}N ) \to \Hom_{A}(M, \iota[S^{-1}N] ) \) needed, with inverse \( \tau^{-1}_{M, S^{-1}N} \) given by \( g \mapsto S^{-1}(g) \). In particular, this shows us that \( \iota \) is a fully faithful functor, as claimed in the problem statement.

To see that \( \iota \) and \( S^{-1} \) are indeed adjoint, let \( f : M \to M' \) be an \(A\)-linear map and \( S^{-1}N\) be an \( S^{-1}A \)-module. We wish to show that the following diagram commutes:

However, this is evident from the definitions, since for any \( S^{-1}A \)-linear \( \alpha : S^{-1}[M] \to S^{-1}N \) and \( m' \in M' \),

As \( \alpha \) and \( m' \) were arbitrary, the diagram commutes. The additional diagram is somewhat trivial, since for \( g : S^{-1}N \to S^{-1}N' \) \(S^{-1}A\)-linear and \( \beta \in \Hom_{S^{-1}A}( S^{-1}[M'], S^{-1}N ) \),

so that the following diagram also commutes:

Solution:

For \(f^i : A^i \to A^{i+1} \), by definition \( \im f^{i} = \ker ( \coker f^i ) \). Now \( \coker f^i \) is a pair \( (B, \coker f^i) \) with \( \coker f^i : A^{i+1} \to B \) universal with respect to the property \( \coker f^i \circ f^{i} = 0 \). Similarly, \( \ker \coker f^i \) is really a pair \( (K, \ker \coker f^i) \) with \( \ker \coker f^i : K \to A^{i+1} \) universal with respect to the property \( \coker f^i \circ \ker \coker f^i = 0 \). Thus, our first complex may be written

For the second sqeuence

By definition, \( H^i := \ker f^i / \im f^{i-1} \), so by universal property of the quotient there is a natural map \( \widetilde{\ker f^i} : H^i \to A^i \) factoring uniquely through \(K\). Similarly, \( \coker f^{i-1} \) is really a pair \( B, \coker f^{i-1} : A^i \to B \) universal with respect to the property that \( \coker f^{i-1} \circ f^{i-1} = 0 \). Now \( f^i : A^i \to A^{i+1} \) induces a map \( \widetilde{f^i} : B \to \im f^i \). Since \( f^{i} \circ f^{i-1} = 0 \), \( f^{i-1} \) factors uniquely through \( K \). If we let \( \pi : A^i \to B \) denote the natural projection, define \( \alpha = \pi \circ \widetilde{\ker f^i} : H^i \to A^i / \im f^{i-1} \), we have \( f^i \circ \textrm{ker}f^i = 0 \) so \( f^i \circ \widetilde{\ker f^i} = 0 \) by a commutivity argument — by passing to the quotient, \( \widetilde{f^i} \circ \alpha = 0\).

Proof:

By definition \( h^i(A^\bullet) = \textrm{nullity} d^i - \textrm{rank}d^{i-1} \) so that

by the rank-nullity theorem.

Proof:

Let us label the morphism \( A^\bullet \to B^\bullet \) as \( \phi = \{ \phi^i \} \), so that the \( \phi^i : A ^i \to B^i \) all satisfy \( g^i \circ \phi^i = \phi^{i+1} \circ f^i \). To see that \( \textrm{Com}_\CC \) is an additive category, we must show three facts:

1. \( \Hom (A^\bullet, B^\bullet) \) is an abelian group.
2. \( \Com{\CC} \) has a zero object.
3. Products exist.

The first statement simply follows from the fact that the category of abelian groups is itself an abelian group, so the infinite direct product exists — thus, we give \( \Hom(A^\bullet, B^\bullet) \) the structure of \( \prod_{i \in \mathscr{I}} \Hom(A^i, B^i) \). The zero object is clearly

Lastly, for each \( i \) take \( P^i = A^i \prod B^i \), \( \pi^i_A : P^i \to A^i \), \( \pi^i_B : P^i \to B^i \) to be the product (exists guaranteed by abelian.) Since we have maps \( d_A^i \circ \pi_A^i : P^i \to A^{i+1} \) and \( d_B^i \circ \pi_B^i : P^i \to B^{i+1} \), by the universal property of the product \( P^{i+1} \) there exists a linear map \(d_P^i : P^i \to P^{i+1} \) that makes

commute. This allows us to consider \( P^\bullet \) as an element in \( \Com{\CC} \) — to see that it is the product simply follows from the fact that any other chain complex with chain maps to \(A^\bullet\) and \(B^\bullet\) would have to uniquely factor through each \( P^i \) on the index level, which by its own respective universal property is unique up to isomorphism. Thus the product exists in \( \Com{\CC} \) so that it is an additive category.

To see that \( \Com{\CC} \) is an abelian category, we need only show that kernels and cokernels exist (as the last two follow from equality on the index level). Let \( \phi : A^\bullet \to B^\bullet \) be a morphism of complexes. Then for each \( i \in \mathscr{I} \), let \( \ker \phi^i : K^i \to A^i \) be the kernel of \(\phi^i\) (which exists as \( \CC \) is abelian). By commutivity, \( \phi^{i+1} \circ d_A^i \circ \ker \phi^i = d_B^{i+1} \circ \phi^i \circ \ker \phi^i = d_B^{i+1} \circ 0 = 0 \), so the universal product of \( \ker \phi^{i+1} \) tells us \(d_A^i \circ \ker \phi \) factors uniquely through the kernel; in other words, there exists a unique morphism \( d_K^i : K^i \to K^{i+1} \) such that \( \ker \phi^{i+1} \circ d_K^i = d_A^i \circ \ker \phi^i \). In addition, since \( \ker \phi^{i+1} \circ d^i_K \circ d^{i-1}_K = d_A^i \circ d_A^{i+1} \circ \ker \phi^{i-1} = 0 \circ \ker \phi = 0\). Since \( \ker \phi^{i+1} \) is a monomorphism, \( d_K^i \circ d_K^{i-1} = 0 \) so that \( K^\bullet \) is indeed a chain complex and \( \ker \phi : K^\bullet \to A^\bullet\) is a morphism of complexes.

In an identical fashion, since \( \CC \) is abelian \( \coker \phi^i \) exists for each \(i \); more precisely, for each \( i \) there exists some \( C^i \) and map \( \coker \phi^i : B^i \to C^i \) universal with respect to the property that \( — \circ \phi = 0 \). Since \( \coker\phi^{i+1} \circ d_B^i \circ \phi^i = \coker \phi^{i+1} \circ \phi^{i+1} \circ d_A^i = 0 \circ d_A^i = 0 \). By the universal property of the coproduct, \( \coker \phi^{i+1} \circ d_B^i \) factors uniquely through \( C^i \), so there exists some morphism \( d_C^i : B^i \to C^i \) such that

commutes. Moreover, \( d_C^i \circ d_C^{i-1} \circ \coker\phi^i = \coker\phi^{i+1} \circ d_B^i \circ d_B^{i-1} = \coker\phi^{i+1} \circ 0 = 0 \) so that \( d_C^i \circ d_C^{i-1} = 0 \). Indeed, this shows that \( C^\bullet \) is an element of \( \Com{\CC} \) and \( \coker\phi \) is a morphisms, such that by a universal argument on the index-level we have that the cokernel exists.

Proof:

Let \( \phi : A^\bullet \to B^\bullet \) be a morphism of complexes

such that \( d_B^i \circ \phi^i = \phi^{i+1} \circ d_A^i \) (where \( d_A^i \) and \( d_B^i \) denote the chain complex maps.) By definition we have \( H^i(A^\bullet) := \ker d_A^i / \im d^{i-1}_A\). Let \( \coker d_B^{i-1} : B^i \to B^i / \im d_B^{i-1} \) denote the natural projection map, and define \( \widetilde{\phi^i} = \coker d_B^{i-1} \circ \phi^i \) — by the universal property of the quotient (which in the context of Vakil, is really inherited from the universal property of the cokernel), \( \widetilde{\phi^i} \) factors uniquely through the quotient \( A^i / \im d^{i-1}_A \) if and only if \( a, a' \in \im d^{i-1}_A \Rightarrow \widetilde{\phi^i}(a) = \widetilde{\phi^i}(a') \).

Suppose \( a, a' \in \im d^{i-1}_A \), so that there exist \( \alpha, \alpha' \in A^{i-1} \) with \( a = d^{i-1}_A (\alpha), a' = d^{i-1}_A(\alpha') \). Then

A similar argument shows \( \widetilde{\phi^i}(\alpha') = 0 \), so that \( \widetilde{\phi^i} \) factors uniquely through \( A^i / \im d^{i-1}_A \). In particular, this induces a natural map \( \overline{\phi^i} : H^i(A^\bullet) \to H^i(B^\bullet) \) such that \( \overline{\phi^i}[s] = [ \phi^i(s') ] \) for any representative \( s' \in [s] \) and this map is well-defined. If we are able to show that \( H^i (\cdot) \) preserves the identity and composition, then we may consider \( H^i(\cdot) \) as a functor \( \Com{\CC} \to \CC \).

The fact that \( H^i \) preserves the identity is fairly obvious; if we let \( id_{A^\bullet} : A^\bullet \to A^\bullet \) denote \( \{ id_{A^i} \} \), then this trivially commutes with the chain map \( d_A^i \) for each \( i\) so that we indeed have \( id_{A^\bullet} \) is a morphism of chain complexes and furthermore the identity (by uniqueness). Then if we choose some arbitrary \( [s] \in H^i(A^\bullet) \) and let \( s' \in [s] \) be a representative, our previous construction gives us

so that by uniqueness of the identity, \( H^i( id_{A^\bullet} ) = id_{H^i(A^\bullet))} \). Similarly, if we let \( \overline{\phi^i} = H^i(\phi) \), \( \overline{\psi^i} = H^i(\psi) \) then

as desired.

Proof:

Suppose

is an exact sequence over some index category \( \mathscr{I} \). This may be "factored" at each \(i\) into short exact sequences

where \( \ker f^j : K^j \to A^j \) denotes the kernel of \( f^j\)/ Since \(\FF \) is both left-exact and right-exact, we get short exact sequence

Now since

is exact if and only if \( (A, f) \) is a kernel of \( g: B \to C \), by an inductive argument we have \( \FF(\ker f^i) = \ker \FF(f^i ) \) for all \( i\). Thus, we may rewrite the above sequence as

Therefore, the sequence

is necessarily exact.

Proof:

1. First, recall from Exercise 1.5.H that for an \(A\)-linear map \( h : M \to N \), we may define an \( S^{-1}A \)-linear map \( \widetilde{h} : S^{-1}M \to S^{-1}N \) by \( \widetilde{h}(\frac{m}{s}) = \frac{h(m)}{s} \). It is easy to see that the map \( S^{-1}(h) = \widetilde{h} \) preserves the identity morphism and (covariantly) preserves composition, so that it is a covariant functor. To see that it is an exact covariant functor, we consider an exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> B @>f>> C @>g>> D @>>> 0 \end{CD} $$

   of \(A\)-modules. Then we must show that \( \widetilde{f} : S^{-1}B \to S^{-1}C \) is injective, \( \widetilde{g} : S^{-1}C \to S^{-1}D \) surjects, and \( \ker \widetilde{g} = \im \widetilde{f} \).

   The fact that \( \widetilde{g} \) is surjective is obvious — take \( \frac{d}{s} \in S^{-1}D \). Since \(g\) surjects, we can find a \(c \in C\) with \(g(c) = d\); thus \( \widetilde{g}( \frac{c}{s} ) = \frac{d}{s} \) by construction. The fact that \( \widetilde{f} \) is injective is also fairly straightforward: suppose \( \widetilde{f}( \frac{b}{s} ) = 0 (= \frac{0}{1}) \). Then \( \frac{f(b)}{s} \sim \frac{0}{1} \) so that there exists some \(a \in S \) with \(a \cdot f(b) = f(ab) = 0\). Since \( f \) is injective, this gives us that \( as = 0 \) — in particular,

   $$ a \cdot (b \cdot 1 - s \cdot 0) = 0 $$

   so that \( \frac{b}{s} \sim \frac{0}{1} \) and thus \( \widetilde{f} \) injective.

   We lastly need to show \( \im \widetilde{f} = \ker \widetilde{g} \) — forward inclusion is easy since we know \( g \circ f = 0 \) so that

   $$ \widetilde{g} \circ \widetilde{f} (\frac{b}{s}) = \widetilde{g} ( \frac{f(b)}{s}) = \frac{g(f(b))}{s} = \frac{0}{s} \sim \frac{0}{1} $$

   for all \( \frac{b}{s} \in S^{-1}B \). To see that \( \ker \widetilde{g} \subseteq \im \widetilde{f} \), fix \( \frac{c}{s} \in \ker \widetilde{g} \). Then there exists some \( a \in S \) such that \( a \cdot g(c) = g(ac) = 0 \) — by exactness of our original sequence\( ac \in \im f \), so there exists some \(b \in B \) with \( f(b) = ac \). In particular,

   $$ \widetilde{f}( \frac{b}{as} ) = \frac{f(b)}{as} = \frac{ac}{as} = \frac{c}{s} $$

   or \( \frac{c}{s} \in \im \widetilde{f} \). Therefore the sequence

   $$ \require{amscd} \begin{CD} 0 @>>> S^{-1}B @>{\widetilde{f}}>> S^{-1}C @>{\widetilde{g}}>> S^{-1}D @>>> 0 \end{CD} $$

   is exact.

2. Refer to the proof of Exercise 1.3.H.

3. From Example 1.2.14 in Vakil and our cursory work on Yoneda's lemma, we know that \( \Hom(M, \cdot) : \Mod{A} \to \textbf{Set} \) is a covariant functor — in fact, since \( \Mod{A} \) is an abelian category, this is really a functor \( \Mod{A} \to Ab \) (the category of abelian groups.) However, each \( \Hom \)-group also has an obvious \( A \)-module structure given by \( a \cdot f := ( x \mapsto a f(x) ) \), so that we get a covariant functor \( \Mod{A} \to \Mod{A} \). To see that it is left-exact, suppose

   $$ \require{amscd} \begin{CD} 0 @>>> B @>f>> C @>g>> D @>>> 0 \end{CD} $$

   is a short exact sequence of \(A\)-modules. We wish to show that the sequence

   $$ \require{amscd} \begin{CD} 0 @>>> \Hom(M, B) @>{f_*}>> \Hom(M, C) @>{g_*}>> \Hom(M, D) \end{CD} $$

   is exact, where \( f_* : \Hom(M, B) \to \Hom( M, C) \) is simply composition on the left by \(f\) (analogous for \(g\)). However, this simply amounts to showing that \( f_* \) is injective and \( \im f_* = \ker g_* \). For the first statement, suppose \( f_*(\beta) = 0_C\) is the zero map \( 0_C : M \to C \). Then for all \( m \in M \), \( f(\beta(m)) = 0 \) — since \( f \) is injective by our original sequence, this implies that \( \beta(m) = 0 \) for all \(m \in M\). In other words, \( \beta \) is necessarily the zero map \( 0_B : M \to B \), so \(f_*\) is injective.

   Next, we wish to show that \( \im f_* = \ker g_* \). Forward inclusion is fairly obvious since for all \( m \in M \),

   $$ g_* ( f_* (\beta))(m) = [g \circ (f \circ \beta)](m) = [(g \circ f) \circ \beta](m) = (0 \circ \beta)(m) = 0 $$

   for any \( \beta \in \Hom(M, B) \) since \( g \circ f = 0 \) in our original diagram. Since \(m\) was arbitrary, \( (g_* \circ f_*)(\beta) \) is the zero map — since \(\beta\) was arbitrary, \( g_* \circ f_* = 0 \) so that \( \im f_* \subset \ker g_* \).

   To see that \( \ker g_* \subset \im f_* \), first realize that the pair \( (B, f) \) necessarily satisfies the universal property of \( \ker g \) from our original short exact sequence. Now suppose \( \gamma : M \to C\) is an element of \( \ker g_* \) so that \( g_* (\gamma) = g \circ \gamma = 0 \). Then by the universal property of the kernel (of \(g\)), \( \gamma \) must uniquely factor through the kernel \( B \) — more precisely, there exists a map \( \iota : M \to B \) with \( f \circ \iota = \gamma \). But this is the same as saying \( f_*(\iota) = \gamma \), so that \( \gamma \in \im f_* \).

4. Similar to part (c) above, we already know that \( \Hom( \cdot, M ) \) is a contravariant functor \( \Mod{A} \to \textbf{Set} \) which may be interpreted as a functor \( \Mod{A} \to \Mod{A} \) since \( \Hom(-, M) \) is an abelian group with an \(A\)-module structure. It suffices to show that this functor is left-exact; suppose we have an exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> B @>f>> C @>g>> D @>>> 0 \end{CD} $$

   We wish to show that the sequence

   $$ \require{amscd} \begin{CD} 0 @>>> \Hom(D, M) @>{g^*}>> \Hom(C, M) @>{f^*}>> \Hom(B, M) \end{CD} $$

   is exact (where \( g^* : \Hom(D, M) \to \Hom(C, M)\) denotes precomposition on the right with \(g\) ); in other words, that \(g^*\) is injective and \( \im g^* = \ker f^* \). For the first assertion, suppose \( \delta : D \to M \) is in the kernel of \( \ker g^* \). Since \(g\) is surjective by our original short exact sequence, for every \( d \in D \) there exists some \( c \in C \) with \( g(c) = d \). Thus, \( \delta(d) = \delta(g(c)) = g^* \delta(c) = 0 \). Since \(d\) was arbitrary, this shows that \( \delta \) is necessarily the zero map.

   Using our usual set-up, it should be fairly immediate that \( \im g^* \subset \ker f^* \) since \( (f^* \circ g^*)(\delta) = \delta \circ (f \circ g) = \delta \circ 0 = 0 \) for each \( \delta : D \to M \), so that \( f^* \circ g^* = 0 \) (composing with zero morphism in abelian category is zero morphism by bilinearity of composition / composing with itself.) To see the other direction, we effectively dualize the argument from part (c). From our axioms of abelian categories, we know that every epimorphism is the cokernel of its kernel. Since \(g\) is surjective and \( \ker g = \im f \) from our original exact sequence, we may simply interpret the pair \( (D, g) \) as the cokernel of \(f\). Now suppose \( \gamma : C \to M \) is in \( \ker f^* \) so that \( f^* \gamma = \gamma \circ f = 0 \).

   

   Then by the universal property of the cokernel \( g \) we must have \( \gamma \) uniquely factors through \( D\); that is, there exists a unique morphism \(\pi : D \to M \) with \( \pi \circ g = \gamma \). But since \( \pi \circ g = g^* \pi \), we have that \( \gamma \in \im g^* \) as desired.

Proof:

By part (a) from the previous exercise, functorality of \( S^{-1} \) gives us a map \( \tau : \Hom_A(M, N) \to \Hom_{S^{-1}A}(S^{-1}M, S^{-1}N) \) defined by \( \tau(f)(m/ s ) = f(m)/s\). For each \(s \in S\), multiplication by \(s\) is an isomorphism in \( \Hom_{S^{-1}A}(S^{-1}M, S^{-1}N) \) so that \( \tau \) factors uniquely through the localization \( S^{-1} \Hom_A(M, N) \) as some \( \widetilde{\tau} \).

Notice that \( \Hom_A(A, N) \cong N \) using the isomorphism \( f \mapsto f(1_A) \). Similarly, for any morphism \( \phi \in \Hom_A(A \oplus A, N) \), we get two morphisms \( \phi_1, \phi_2 : A \to N \) given by \( \phi_1(a) = \phi(a, 0) \) and \( \phi_2 (a) = \phi(0, a) \). There is an obvious isomorphism \( \Hom_A(A \oplus A, N) \cong \Hom_A(A, N) \oplus \Hom_A(A, N) \) given by \( \phi \mapsto (\phi_1, \phi_2 ) \). Proceeding inductively, it is easy to see that \( \Hom_A (A^{oplus m}, N) \cong N^{\oplus m} \) for any finite \(m\). Passing to the localisation, Exercise 1.3.F(b) tells us

so that \( S^{-1}\Hom_A(M, N) \) and \( \Hom_{S^{-1}A}(S^{-1} M, S^{-1}N) \) are isomorphic for \(M = A^{\oplus m} \). This isomorphism \( \tau_m \) can also be explicitly constructed by fixing a basis \( a_1, \dots, a_n \) for \( A^{\oplus n} \), noticing \( \frac{a_1}{1}, \dots, \frac{a_n}{1} \) is a basis for \( S^{-1}(A^{\oplus m}) \) as an \( S^{-1}A \)-module, and showing that the map \( f \mapsto \widetilde{f} \) where \( \widetilde{f}( \frac{a_i}{s} ) = \frac{ f(a_i) }{s} \) is a well-defined isomorphism.

Since \( \Hom_A( —, N ) \) and \( \Hom_{S^{-1}A}( — , S^{-1}N) \) are left-exact contravariant functors by the previous exercise, we get exact sequences

and

Along with our isomorphisms described above, we get a commuting diagram

By using the five lemma from section 1.7, it follows that \( \widetilde{\tau}\) is an isomorphism.

Proof:

1. Recall that a sequence

   $$ \require{amscd} \begin{CD} A @>{f}>> B @>{g}>> C @>>> 0 \end{CD} $$

   is exact if and only if \( C \) is a cokernel of \(f\). Consider the cokernel of \(d^i\) given by some object \(D\) and map \( \coker d^i : C^{i+1} \to D\). Since cokernels are naturally epimorphisms, there is an exact sequence

   $$ \require{amscd} \begin{CD} C^i @>{d^i}>> C^{i+1} @>{\coker d^i}>> D @>>> 0 \end{CD} $$

   Since \( \FF \) is a right-exact functor by supposition, the sequence

   $$ \require{amscd} \begin{CD} \FF(C^i) @>{\FF(d^i)}>> \FF(C^{i+1}) @>{ \FF(\coker d^i) }>> \FF(D) @>>> 0 \end{CD} $$

   is exact. But then by our first statement we must have \( \FF( \coker d^i) \) is a cokernel of \( d^i \) (up to isomorphism) — in other words, we have \( \FF(\coker d^i) \cong \coker \FF(d^i)\).

   Next, since \( C^\bullet \) and \( \FF C^\bullet \) are chain complexes, we have short exact sequences

   $$ \require{amscd} \begin{CD} 0 @>>> \im d^i @>{ \ker \kappa }>> C^{i+1} @>{\kappa}>> \coker d^i @>>> 0 \end{CD} $$

   and

   $$ \require{amscd} \begin{CD} 0 @>>> \im \FF(d^i) @>{ \ker \gamma }>> \FF(C^{i+1}) @>{\gamma}>> \coker \FF(d^i) @>>> 0 \end{CD} $$

   It is important to recognize here that the image is defined (in Vakil) to be the kernel of the cokernel, hence the notation above. Let \( \tau \) denote the isomorphism \( \FF(\coker d^i) \cong \coker \FF(d^i) \); applying the right-exact functor \( \FF \) to the first short exact sequence, we get a commuting diagram

   $$ \require{amscd} \begin{CD} @. \FF(\im d^i) @>{ \FF(\ker \kappa) }>> \FF(C^{i+1}) @>{ \FF(\kappa) }>> \FF( \coker d^i ) @>>> 0 \\ @. @. @| @V{\tau}VV \\ 0 @>>> \im \FF(d^i) @>{\ker \gamma}>> \FF(C^{i+1}) @>{\gamma}>> \coker \FF(d^i) @>>> 0 \end{CD} $$

   By commutivity of the right-half of the diagram, \( \gamma \circ \FF(\ker \kappa) = \tau \circ \FF( \kappa \circ \ker \kappa ) = \tau \circ 0 = 0 \). By the universal property of the kernel, there exists a unique map \( \psi: \FF( \im d^i ) \to \im \FF(d^i) \) which makes the diagram commute. (We will not prove \( \psi \) is surjective as it is not needed. )

   Lastly, consider the exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> H^i(C^\bullet) @>{\ker \widetilde{d^i} }>> \coker d^{i-1} @>{ \widetilde{d^i} }>> \im d^i @>>> 0 \end{CD} $$

   from Exercise 1.6.A, where \( \widetilde{d^{i}} \) is the natural induced map from the universal property of the cokernel (since \( d^i \circ d^{i-1} = 0 \) by definition of a chain complex). Again, it is natural to consider \( H^i(C^\bullet) \) as the kernel of this induced map. Since \( \FF(C^\bullet) \) is also a chain complex, we get a similar sequence

   $$ \require{amscd} \begin{CD} 0 @>>> H^{i}(\FF C^\bullet) @>{\ker \widetilde{ \FF(d^i) } }>> \coker \FF( d^{i-1} ) @>{ \widetilde{\FF(d^i) } }>> \im \FF(d^i) @>>> 0 \end{CD} $$

   where, again, \( \widetilde{ \FF(d^i) } \) denotes the induced map from the universal property of the cokernel. Notice we now have a diagram

   $$ \require{amscd} \begin{CD} \FF \coker (d^{i-1}) @>{ \FF (\widetilde{d^i}) }>> \FF \im(d^i) @>{ \FF (\ker \kappa) }>> \FF(C^{i+1}) \\ @| @V{\psi}VV @| \\ \coker \FF(d^{i-1}) @>{ \widetilde{\FF(d^i)} }>> \im \FF(d^i) @>{\ker \gamma}>> \FF(C^{i+1}) \end{CD} $$

   Although the rows need not be exact, we know that the outside rectangle necessarily commutes as well as the right square. Since \( \ker \gamma \) is a monomorphism, any two paths from \( \FF \coker(d^{i-1}) \) to \( \FF(C^{i+1}) \) must be the same. Thus, the left square must commute as well. Using this simply fact, as well as applying our right-exact functor \( \FF \) to the exact sequence above, we get a commuting diagram

   $$ \require{amscd} \begin{CD} \FF H^i(C^\bullet) @>{ \FF( \ker \widetilde{d^i} ) }>> \FF \coker (d^{i-1}) @>{ \FF (\widetilde{d^i}) }>> \FF \im(d^i) @>>> 0 \\ @. @| @V{\psi}VV\\ H^i( \FF C^\bullet ) @>{\ker \widetilde{\FF(d^i)} }>> \coker \FF(d^{i-1}) @>{ \widetilde{\FF(d^i)} }>> \im \FF(d^i) @>>> 0 \end{CD} $$

   As \( \FF (\widetilde{d^i}) \circ \FF( \ker \widetilde{d^i} ) = \FF( \widetilde{d^i} \circ \ker \widetilde{d^i} ) = \FF(0) = 0 \) and the diagram commutes, we must also have that \( \widetilde{ \FF(d^i) } \circ \FF( \ker \widetilde{d^i} ) = 0 \) so that \( \FF( \ker \widetilde{d^i} ) \) uniquely factors through the kernel of \( \widetilde{ \FF(d^i) } \) — in other words, there exists a unique morphism \( \FF H^i(C^\bullet) \to H^i( \FF C^\bullet ) \) as desired.

2. Dually, recall a sequence

   $$ \require{amscd} \begin{CD} 0 @>>> A @>f>> B @>g>> C \end{CD} $$

   is exact if and only if \(f\) is a kernel of \(g\). Now for any complex \( C^\bullet \) (need not be exact) we get

   $$ \require{amscd} \begin{CD} 0 @>>> \ker d^i @>>> C^i @>{d^i}>> \im d^i @>>> 0 \end{CD} $$

   is an exact sequence. Applying our left-exact functor \( \FF \), we get the two exact sequences

   $$ \require{amscd} \begin{CD} 0 @>>> \FF( \ker d^i ) @>>> \FF C^i @>{ \FF(d^i) }>> \FF( \im d^i )\\ @. @. @. \\ 0 @>>> \ker \FF(d^i) @>>> \FF C^i @>>> \im \FF(d^i) @>>> 0 \end{CD} $$

   As before, since the top sequence is exact, we have an isomorphism \( \FF(\ker d^i) \cong \ker \FF(d^i) \). Additionally by interpreting the image as the kernel of the cokernel, we get a unique morphism \( \phi : \FF( \im d^i ) \to \im \FF(d^i) \). Using the exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> \im d^{i-1} @>>> \ker d^i @>>> H^i(A^\bullet) @>>> 0 \end{CD} $$

   and applying the same derivations as before, diagram chasing the complex

   $$ \require{amscd} \begin{CD} 0 @>>> \FF( \im d^{i-1}) @>>> \FF(\ker d^i) @>>> \FF H^i(C^\bullet) \\ @. @VVV @| \\ 0 @>>> \im \FF( d^{i-1}) @>>> \ker \FF d^i @>>> H^i \FF C^\bullet @>>> 0 \end{CD} $$

   and interpreting the cohomology \( H^i \FF C^\bullet \) as the cokernel of the induced map \( \im \FF d^{i-1} \to \ker \FF d^i \) yields a unique morphism \( H^i \FF C^\bullet \to \FF H^i(C^\bullet) \).

   ^ May need to further expand this at a later date, but am postponing for the moment. ^



3. It is in fact superfluous to show that the two induced morphisms above are inverses; we know that a right-exact functor commutes with cokernels and a left-exact functor commutes with kernels. Since the image of a morphism is defined as the kernel of its cokernel, it should be clear that an exact functor (that is, left and right exact functor) commutes with images. Now recall that the quotient \( B / A \) is defined to be the cokernel of a monomorphism \( A \hookrightarrow B \). In other words, an exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> A @>>> B @>>> C @>>> 0 \end{CD} $$

   tells us that \( C \cong B / A \). Using the exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> \im d^{i-1} @>>> \ker d^i @>>> H^i (C^\bullet) @>>> 0 \end{CD} $$

   since our functor \( \FF \) is exact, we get a commutative diagram

   $$ \require{amscd} \begin{CD} 0 @>>> \FF( \im d^{i-1} ) @>>> \FF( \ker d^i ) @>>> \FF H^i (C^\bullet) @>>> 0\\ @. @| @| \\ 0 @>>> \im \FF(d^{i-1}) @>>> \ker \FF(d^i) @>>> H^i( \FF C^\bullet ) @>>> 0 \end{CD} $$

   (where the vertical equals signs really mean "up to isomorphism"). But then we get an induced exact sequence

   $$ \require{amscd} \begin{CD} 0 @>>> \im \FF( d^{i-1} ) @>>> \ker \FF( d^i ) @>>> \FF H^i(C^\bullet) @>>> 0 \end{CD} $$

   so that by the fact noted above, \( \FF H^i( C^\bullet ) \cong H^i ( \FF C^\bullet ) \).

Proof:

Suppose we let \( \kappa_i : \ker h_i \to A_i \) denote the kernel of \( h_i \) for each \( i \in \II \) — our first objective is to show that there are unique morphisms \( \ker h_i \to \ker h_j \) for any \( i \to j \) in \( \II \) (which would then make the kernel maps \(\kappa_j \) natural morphisms). By commutivity,

so that by the universal property of the kernel, there is a unique morphism \( c(i \to j) : \ker h_i \to \ker h_j \) that makes the following diagram commute

Recall that natural transformations factor through limits by the universal property of limits. More specifically, assuming that our \( \varprojlim A_i \) and \( \varprojlim B_i \) exist by supposition, and letting \( \phi_j : \varprojlim A_i \to A_j \) denote the canonical morphisms, then for each \( j \) we have maps \( h_j \circ \phi_j : \varprojlim A_i \to B_j \) which necessarily commute with \( b(i \to j) \) (as \( h \) is a natural transformation). By the universal property of \( \varprojlim B_i \), there exists some unique morphism \( \phi : \varprojlim A_i \to \varprojlim B_i \)

By supposition we have that \( \varprojlim \ker h_i \) exists with canonical maps \( \xi_j : \varprojlim \ker h_i \to \ker h_j \) for each \( j \in \II \). But then by a similar logic to the paragraph above, for each \( j \) we get maps \( \kappa_j \circ \xi_j : \varprojlim \ker h_i \to A_j \) which necessarily commute with the transitions \( a(i \to j) \), so that there is a unique map \( \kappa : \varprojlim \ker h_i \to \varprojlim A_i \) making the following diagram commute

We wish to show that \( \varprojlim \ker h_i \) satisfies the universal property of the kernel of \( h \). First, by commutivity it is easy to see that

If \( M \) is any other object in \( \CC \) with morphism \( \mu : M \to \varprojlim A_i \) such that \( h \circ \mu = 0 \), then by commutivity for each \( j \in \II \)

so that \( \phi_j \circ \mu \) factors uniquely through \( \ker h_j \) yielding a unique morphism \( \nu_j : M \to \ker h_j \) that commutes with the necessary maps. By the universal property of the limit \( \varprojlim \ker h_i \), there exists another unique morphism \( M \to \varprojlim \ker h_i \) which then must necessarily commute with \( \kappa \) (by commutivity of basically everything else). Thus, \( \varprojlim \ker h_i \) satisfies the universal property of the kernel of \( h \), so that the two are necessarily isomorphic.

Solution:

Let \( \CC \) be an arbitrary category. In order for double limits to even have a source, we must have two (not necessarily distinct) index categories \( I \) and \(J\). To be as precise as possible (at least at the level of, say, a graduate student), I find it appropriate to define the category of cones over \( \CC \) indexed by \(I\), which we will denote \( \cone{\CC}{I} \). An object of \( \cone{\CC}{I} \) is a triple \( (C, \FF, \phi) \), where \( C \) is an object in \( \CC \), \( \FF : I \to \CC \) is a functor / diagram, and \( \phi \) is a family of morphisms \( \phi_i : N \to \FF(i) \) that commute with maps \( \FF( i \to i') \). A morphism of cones \( \tau : (C, \FF, \phi) \to (D, \GG, \psi) \) is then a map \( \tau_0 : C \to D \) and a collection of natural transformations \( \widetilde{\tau} : \FF \Rightarrow \GG \) making the resulting triangular prism diagram commute. It should be clear that a limit is simply a universal object in the category of cones.

Consider the expression \( \varprojlim_i \varprojlim_j A_{i,j} \) where \( i \) ranges over \(I\) and \(j\) ranges over \( J \) — in order for this to make sense fully, we should in fact think of the \( \varprojlim_i \) as a functor \( \cone{\CC}{J} \to \cone{Cone_\CC}{I} \). For each \( i \in I \), let \( B_i = \varprojlim_j A_{i,j} \) denote the limit interpreted as a cone. Let \( S = \varprojlim_i B_i \), and fix some \( j_0 \in J \). For each \(i \in I \), we have maps \( \pi_i : S \to B_i \) commuting with morphisms in \( \cone{\CC}{J} \). Since \( B_i \) is also a cone indexed by \( J\), there exists a morphism \( \phi_{j_0}^i : B_i \to A_{i,j_0} \) which commutes with maps \( A_{i, j_0} \to A_{i, j_1} \) (induced from some \( j_0 \to j_1 \)). Composing, we get maps \( \phi_{j_0}^i \circ \pi_i : S \to A \) which commute with morphisms induced from \( i \to i' \) by construction of cone morphisms (c.f. "commuting triangular prism" above ). By the universal property of limits, there is necessarily a map \( \xi_{j_0} : S \to \varprojlim_i A_{i, j_0} \) which commutes in the sense that if \( \psi^{j_0}_{i_0} : \varprojlim_i A_{i, j_0} \to A_{i_0, j_0} \) denotes our canonical map at index \(i_0 \in I\), then \( \psi_{i_0}^{j_0} \circ \xi_{j_0} = \phi^{j_0}_{i_0} \circ \pi_{i_0} \).

Since \( j_0 \in J \) was fixed above, we may now allow it to range over \( J\) so that we get a family of maps \( \xi_j \) from \(S\) to our limits \( \varprojlim_i A_{i,j} \). If \( \tau' \) is any cone morphism \( \varprojlim A_{i, j_0} \to \varprojlim A_{i, j_1} \), then the fact that our \( \xi_j \) commute with \( \pi_i \) which commute with cone morphisms implies that the maps \( S \to \varprojlim_i A_{i, j} \) commute with the necessary morphisms — by the universal property of the limit, there then exists a map \( S \to \varprojlim_j \varprojlim_i A_{i, j} \).

In order to show that the natural map above is in fact an isomorphism, one can show that any other \(S'\) with maps to each \( \varprojlim_{i} A_{i,j} \) must factor through \(S\) by working backwards and building maps \( S' \to B_j \) which invoke the universal property of the limit for \(S\). Alternatively, one could just mimic the exact same argument above starting with \(\varprojlim_j \varprojlim_i A_{i,j} \) this time, and get a map \( S \leftarrow \varprojlim_j \varprojlim_i A_{i, j} \) which would necessarily be an inverse. Both techniques seem messy and will be left for later.

Proof:

Let \( \II \) be a filtered index category, and \( A, B, C : \II \to \Mod{A} \) be functors. Furthermore, let \( f : A \Rightarrow B \) and \( g : B \rightarrow C \) be natural transformations so that, for each \(i\)

is exact. From Exercise 1.4.E, we know that \( \varinjlim A_i, \varinjlim B_i \) and \( \varinjlim C_i \) exist — let the canonical maps be \( \alpha_j : \varinjlim A_i \to A_j, \beta_j : \varinjlim B_i \to B_j\) and \( \gamma_j : \varinjlim C_i \to C_j \). Additionally, we know that \( \varprojlim \) is right-exact as described above — thus, we need only show that the induced map \( \widetilde{f} : \varinjlim A_i \to \varinjlim B_i \) (arising from the universal property of the colimit, since we get maps from each \( A_i \) to \( \varinjlim B_i \) via composition ) is injective.

Suppose \( x = (x_i, i) \in \ker \widetilde{f} \) so that \( f(x) = 0 \) in \( \varinjlim B_i \) — from Vakil's discussion preceding Exercise 1.4.E, this implies that the \( f_i(x_i) \) are eventually \(0\) in the diagram. Thus, we can find some \( j \in \II \) with \( f_j(x_j) = 0 \) in \( B_j \) and \( f_k(x_k) = 0 \) for any succeeding \( j \to k \). As the sequence

is exact by supposition, we must have that \( x_j = 0 \) (and also \( x_k = 0 \) for any succeeding \( j \to k \)). But then clearly our \( (x_i, i) \) are "eventually zero in the diagram" so that \( (x_i, i) \sim 0 \) and thus the filtered colimit in \( \Mod{A} \) is exact.

Proof: