Homological Algebra Problem Sets

For \(R\) a PID, show that an \(R{\hbox{-}}\)module \(A\) is divisible if and only if \(A\) is injective.

Recall that a module is divisible if and only if for every \(r\neq 0 \in R\) and every \(a\in A\), we have \(a=br\) for some \(b\in A\).

Note: we’ll assume \(R\) is commutative, and since \(R\) is a domain, it has no nonzero zero divisors and thus all elements \(r\in R\) are left-cancelable.

\(\implies\): Suppose \(A\) is divisible, we then want to show every \(R{\hbox{-}}\)module morphism of the following form lifts, where we regard the ideal \(J\) and the ring \(R\) as \(R{\hbox{-}}\)modules:

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Since \(R\) is a PID, we have \(J = jR\) for some \(j\in R\), so it suffices to produce lifts of the following form:

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Consider \(f(j)\in A\). Since \(A\) is divisible, we have \(A = jA\), so we can write \(f(j) = j \mathbf{a}'\) for some \(\mathbf{a}' \in A\). Using \(R{\hbox{-}}\)linearity and the fact that \(j\) is left-cancelable, we have \begin{align*} jf(1_R) = f(j) = j\mathbf{a}' \implies f(1_R) = \mathbf{a}' .\end{align*}

Thus we can set \begin{align*} \tilde f: R &\to A\\ 1_R &\mapsto \mathbf{a}' ,\end{align*} and extending \(R{\hbox{-}}\)linearly yields a well-defined \(R{\hbox{-}}\)module morphism. Moreover, the diagram commutes by construction, since \(\iota(1_R) = 1_R\).


\(\impliedby\): Suppose $A\in {}_{R}{\mathsf{Mod}} $ is injective, where by Baer’s criterion we equivalently have a lift of the following form for every \(J{~\trianglelefteq~}R\):

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Let \(j\in R\) be a nonzero element that is not a zero-divisor, we then want to show that \(A = jA\), i.e. that for every \(\mathbf{a} \in A\), there is a \(\mathbf{a}' \in A\) such that \(\mathbf{a} = j \mathbf{a}'\). Fixing \(\mathbf{a}\in A\), define a map \(f_a: J\to A\) in the following way: for \(x\in J\), use the fact that \(\left\langle{ j }\right\rangle\coloneqq jR\) to first write \(x = jr\) for some \(r\in R\), and then set \(f_a(x) = f_a(jr) \coloneqq r \mathbf{a}\). To summarize, we have \begin{align*} f_a: J = jR &\to R \\ x = jr &\mapsto r\mathbf{a} .\end{align*} By injectivity, we can take the inclusion \(jR\hookrightarrow R\) and get a lift:

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We can now use the fact that \begin{align*} r \mathbf{a} &= f_a(jr) \\ &= \tilde f_a(\iota(jr)) \\ &= \tilde f_a(jr) \\ &= jr \tilde f_a(1_R) && \text{using $R{\hbox{-}}$linearity and }j,r\in R \\ &= rj \tilde f_a(1_R) && \text{since $R$ is commutative} \\ \implies \mathbf{a} &= j\tilde f_a(1_R) \in j A ,\end{align*} where in the last step we have canceled an \(r\) on the left. So in the definition of divisibility, we can take \begin{align*} \mathbf{a}' \coloneqq\tilde f_a(1_R) ,\end{align*} and letting \(\mathbf{a}\) range over all elements of \(A\) yields the desired result.

Calculate \(\operatorname{Ext} _{\mathbf{Z}}^i({\mathbf{Z}}/p, {\mathbf{Z}}/q)\) for distinct primes \(p, q\).

The following are several claims that are later used in the actual solution:

For any \(m\in {\mathbf{Z}}\), \begin{align*} \mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}/n, {\mathbf{Q}}/{\mathbf{Z}}) \cong {\mathbf{Z}}/n .\end{align*}

Note that there is an injection \begin{align*} 1 \to \mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}/n, {\mathbf{Q}}/{\mathbf{Z}}) \hookrightarrow\mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}, {\mathbf{Q}}/{\mathbf{Z}}) ,\end{align*} which follows from the fact that there is a SES \begin{align*} 1 \to {\mathbf{Z}}\xrightarrow{x\mapsto nx} {\mathbf{Z}}\xrightarrow{\pi_n} {\mathbf{Z}}/n \to 1 \end{align*} where \(\pi_m\) is the canonical quotient morphism, and applying the left-exact contravariant functor \(\mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}/n, {\mathbf{Q}}/{\mathbf{Z}})\) yields the first exact sequence above. We use this to identify the former as a submodule of the latter, and note that for any \({\mathbf{Z}}{\hbox{-}}\)module morphism \({\mathbf{Z}}\xrightarrow{f} {\mathbf{Q}}/{\mathbf{Z}}\),

  • Since \({\mathbf{Z}}\) is a free \({\mathbf{Z}}{\hbox{-}}\)module with generator 1, \(f\) is entirely determined by \(f(1)\), and

  • \(f\) descends to a map \(\tilde f: {\mathbf{Z}}/n \to {\mathbf{Q}}/{\mathbf{Z}}\) if and only if \(f(n) \in {\mathbf{Z}}\), i.e. \(f(n) = [0]\) is in the equivalence class of zero in the quotient, and so \begin{align*} [1] = [0] = f(n) = nf(1) .\end{align*}

Using this injection, we can identify the submodule \(\mathop{\mathrm{Hom}}({\mathbf{Z}}/n, {\mathbf{Q}}/{\mathbf{Z}})\) as all of those morphism \({\mathbf{Z}}\to {\mathbf{Q}}/{\mathbf{Z}}\) which descend to make the following diagram commute.

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To characterize these, it suffices to determine all of the possible images \(f(1)\). Moreover, we can restrict our attention to coset representatives in the interval \([0, 1) \cap{\mathbf{Q}}\subseteq {\mathbf{R}}\), where we want to find all \(q \coloneqq f(1) \in [0, 1)\) such that \(nq = 1\). A complete list of \(n\) such representatives is given by \begin{align*} q \in \left\{{ 0, {1\over n}, {2\over n}, \cdots, {n-1 \over n} }\right\} .\end{align*}

Setting \(f_i(1) \coloneqq\left[ {i\over n}\right]\) (where we take the equivalence class mod \({\mathbf{Z}}\)) yields \(n\) distinct morphisms \(f_i: {\mathbf{Z}}\to {\mathbf{Q}}/{\mathbf{Z}}\) that descend to \(\tilde f_i: {\mathbf{Z}}/n \to {\mathbf{Q}}/{\mathbf{Z}}\). We can define a map \begin{align*} \Psi: {\mathbf{Z}}&\to \mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}/n, {\mathbf{Q}}/{\mathbf{Z}}) \\ i &\mapsto f_i ,\end{align*} and using the fact that if \(i = i' \operatorname{mod}n\), write \(i' = i + kn\) for some \(k\in {\mathbf{Z}}\), then \begin{align*} f_{i'}(1) = f_{i + kn}(1) = \left[ {i + kn \over n}\right] = \left[{i\over n} + k\right] = \left[ {i\over n} \right] = f_{i}(1) ,\end{align*} since \(k\in {\mathbf{Z}}\), so by the first isomorphism theorem \(\Psi\) descends to an isomorphism \begin{align*} \tilde \Psi: {\mathbf{Z}}/n \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}/n, {\mathbf{Q}}/{\mathbf{Z}}) .\end{align*}

\({\mathbf{Q}}/{\mathbf{Z}}\) is an injective object in \({\mathbf{Z}}{\hbox{-}}\)modules.

By the previous exercise, it suffices to show that \({\mathbf{Q}}/{\mathbf{Z}}\) is divisible. More generally, if any group \(G\) is divisible and \(N{~\trianglelefteq~}G\) is a normal subgroup, then \(G/N\) will be divisible. This follows from the fact that if \(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mub\mkern-1.5mu}\mkern 1.5mu \in G/N\) and \(n\in {\mathbf{Z}}\), we can write \(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu = a + N\) and \(\mkern 1.5mu\overline{\mkern-1.5mub\mkern-1.5mu}\mkern 1.5mu = b + N\) for some coset representatives, use divisibility to write \(a = nb\), and then compute \begin{align*} \mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu = a + N = (nb) + N \coloneqq n(b + N) = n \mkern 1.5mu\overline{\mkern-1.5mub\mkern-1.5mu}\mkern 1.5mu .\end{align*}

That \({\mathbf{Q}}\) is divisible is a straightforward check: let \(n\in {\mathbf{Z}}\) and \(a\in {\mathbf{Q}}\), we then want a \(b\in {\mathbf{Q}}\) such that \(a = nb\), and \(b \coloneqq{a\over n} \in {\mathbf{Q}}\) works. Since \({\mathbf{Q}}\) is an abelian group, \({\mathbf{Z}}\) is automatically normal, and the result follows.

\begin{align*} {{\mathbf{Z}}/n \over m({\mathbf{Z}}/n)} \cong {\mathbf{Z}}/d && d \coloneqq\gcd({\mathbf{Z}}/m, {\mathbf{Z}}/n) .\end{align*}

Using \begin{align*} M\otimes_R {A\over I} \cong {M\over IM} \in {}_{R}{\mathsf{Mod}} ,\end{align*} and taking

  • \(M \coloneqq{\mathbf{Z}}/m\),
  • \(A \coloneqq{\mathbf{Z}}\),
  • \(I \coloneqq n{\mathbf{Z}}\),

we have \begin{align*} {\mathbf{Z}}/m \otimes_{\mathbf{Z}}{\mathbf{Z}}/n \cong {{\mathbf{Z}}/m \over n\qty{{\mathbf{Z}}/m} } && \in {}_{{\mathbf{Z}}}{\mathsf{Mod}} .\end{align*}

We can now use the map \begin{align*} \phi: {\mathbf{Z}}&\to {\mathbf{Z}}/m \otimes_{\mathbf{Z}}{\mathbf{Z}}/n \\ x & \mapsto x(1 \otimes 1) \end{align*} and compute \begin{align*} \ker \phi &= \left\{{x\in {\mathbf{Z}}{~\mathrel{\Big\vert}~}x(1\otimes 1) = 0}\right\}\\ &= \left\{{x\in {\mathbf{Z}}{~\mathrel{\Big\vert}~}n\divides x \text{ or } m \divides x}\right\}\\ &= \left\langle{ n, m }\right\rangle \\ &= \left\langle{ \gcd(n, m) }\right\rangle && \text{by Bezout's theorem} \\ &\coloneqq\left\langle{ d }\right\rangle .\end{align*} Now applying the first isomorphism theorem yields the result.

We’ll follow the procedure outlined in Weibel:

  • Define the contravariant functor \(F({-}) \coloneqq\mathop{\mathrm{Hom}}_{\mathbf{Z}}({\mathbf{Z}}/p, {-})\), then noting that it is left-exact, it has right-derived functors.
  • Find an injective resolution \(I\) of \({\mathbf{Z}}/q\).
  • Write \(F(I)\) as a new (not necessarily exact) chain complex.
  • Compute \(\operatorname{Ext} _{\mathbf{Z}}^i({\mathbf{Z}}/p, {\mathbf{Z}}/q) \coloneqq R^i F({\mathbf{Z}}/q) \coloneqq H^i(F({\mathbf{Z}}/q))\).

We can first take the following injective resolution:

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This is a chain complex by construction, since \(d^2([1]_q) = \left[ q \qty{1\over q} \right] = [1] = [0]\). We now delete the augmentation and apply \(F({-})\):

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Here we immediately simplify by applying the isomorphism from the earlier claim. Noting that \(d^0(x) \coloneqq qx\) was multiplication by \(q\), we have \({{\partial}}^0(f) = d^0 \circ f\) is post-composition by the multiplication by \(q\) map, and \(\tilde {{\partial}}^0\) similarly becomes multiplication by \(q\).

We now take homology: \begin{align*} \operatorname{Ext} ^1_{{\mathbf{Z}}}({\mathbf{Z}}/p, {\mathbf{Z}}/q) \coloneqq R^1 F({\mathbf{Z}}/q) \coloneqq{\ker {{\partial}}^1 \over \operatorname{im}{{\partial}}^0} = {{\mathbf{Z}}/p \over q \qty{{\mathbf{Z}}/p}} \cong {\mathbf{Z}}/ d{\mathbf{Z}}\cong 1 ,\end{align*} where \(d \coloneqq\gcd(p, q) = 1\) if \(p, q\) are coprime.

Let \(A\in {\mathsf{Ab}}\), and show that the following map is injective: \begin{align*} {\varepsilon}_A : A &\to I(A) \coloneqq \prod_{f\in \mathop{\mathrm{Hom}}_{\mathsf{Ab}}(A, {\mathbf{Q}}/{\mathbf{Z}}) } {\mathbf{Q}}/{\mathbf{Z}}\\ a &\mapsto \mathbf{a} \text{ where } \mathbf{a}(f) \coloneqq f(a) \in {\mathbf{Q}}/{\mathbf{Z}} ,\end{align*} i.e. when looking at the image \({\varepsilon}_A(a)\) in the product, the component indexed by \(f\) is an element of \({\mathbf{Q}}/{\mathbf{Z}}\) obtained by evaluating \(f(a)\).

Hint: if \(a\in A\), find a map \(f: a {\mathbf{Z}}\to {\mathbf{Q}}/{\mathbf{Z}}\) with \(f(a) \neq 0\) and extend this to a map \(f': A\to {\mathbf{Q}}/{\mathbf{Z}}\).

By contrapositive, we’ll suppose \(a\neq 0\) and show \({\varepsilon}_A(a) \neq 0\). Following the hint, we first consider the cyclic subgroup \(a{\mathbf{Z}}= \left\{{an {~\mathrel{\Big\vert}~}n\in {\mathbf{Z}}}\right\}\) and define a map \begin{align*} f_a: a{\mathbf{Z}}&\to {\mathbf{Z}}\\ an & \mapsto n .\end{align*}

We now pick \(\ell > 1 \in {\mathbf{Z}}\) to be any integer, and define a composition \(f: a{\mathbf{Z}}\to {\mathbf{Q}}/{\mathbf{Z}}\):

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By choice of \(\ell\), this map satisfies \(f(a) = [1/\ell] \neq 0\), so the map is nonzero. Since \({\mathbf{Q}}/{\mathbf{Z}}\) is injective, the universal property provides a lift \(\tilde f\):

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Since \(\tilde f\) lifts \(f\), it is also nonzero. But now we can check that \begin{align*} {\varepsilon}_A(a)(f) \coloneqq f(a) \neq 0 ,\end{align*} so the \(f\) component of the image of \(a\) is nonzero and thus \(\mathbf{a} \coloneqq{\varepsilon}_A(a) \neq 0\) in the product.

Sunday, February 28th

If \(U: \mathsf{B} \to \mathsf{C}\) is right-exact functor, show that \begin{align*} U(L_i F) \cong L_i(UF) .\end{align*}

We’ll show that \((U \circ L_i F)(X) \cong (L_i(U\circ F) )(X)\) for every object \(X\). Starting with the left-hand side, to compute left-derived functors, we’ll need projective resolutions, so let \(P \to X\) be a projective resolution of \(X\). Fixing labeling, we have the following situation:

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We now have by definition \begin{align*} L_iF(X) \coloneqq{ \ker F({{\partial}}_i) \over \operatorname{im}F({{\partial}}_{i+1})} \implies U\qty{ L_iF(X)} \coloneqq U\qty{ \ker F({{\partial}}_i) \over \operatorname{im}F({{\partial}}_{i+1})} .\end{align*}

For the right-hand side, we can take the same projective resolution \(P \to X\), and apply a similar process:

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Again, by definition, \begin{align*} (L_i(UF))(X) \coloneqq { \ker (UF)({{\partial}}_i) \over \operatorname{im}(UF)({{\partial}}_{i+1})} ,\end{align*}

and thus it suffices to show that there is an isomorphism \begin{align*} U\qty{ \ker F({{\partial}}_i) \over \operatorname{im}F({{\partial}}_{i+1})} \xrightarrow{\sim} { \ker (UF)({{\partial}}_i) \over \operatorname{im}(UF)({{\partial}}_{i+1})} .\end{align*}

To show this, we apply the exact functor \(U\) to the following SES to produce a new SES, from which we’ll produce the desired isomorphism \(f\):

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Here \(\tilde \pi_i\) is the natural quotient map, whose image is \(\operatorname{coker}U(\iota_i)\). Finally, the map \(f\) exists an any abelian category, using that whenever \(0\to A \xrightarrow{g_1} B \xrightarrow{g_2} C\to 0\) is exact, there is an isomorphism \(C \xrightarrow{\sim} B/\operatorname{im}(g_1)\).

\envlist
  • If \(0\to M \to P \to A \to 0\) is exact with \(P\) projective or \(F{\hbox{-}}\)acyclic, show that \begin{align*} L_i F(A) \cong L_{i-1}FM && i\geq 2 .\end{align*}

  • Show that if \begin{align*} 0 \to M_m \to P_m \to P_{m-1} \to \cdots \to P_0 \to A \to 0 \end{align*} is exact with \(P_i\) projective or \(F{\hbox{-}}\)acyclic, then \begin{align*} L_iF(A) \cong L_{i-m-1}F(M_m) && i\geq m+2 .\end{align*}

    • Moreover show that \(L_{m+1} F(A)\) is the kernel of \(F(M_m) \to F(P_m)\).
  • Conclude that if \(P\to A\) is an \(F{\hbox{-}}\)acyclic resolution of \(A\), then \(L_i F(A) = H_i(F(P))\).

\envlist

\begin{align*} L_i F(A) \cong L_{i-1}FM && i\geq 2 .\end{align*}

Following the proof of Weibel Theorem 2.4.6, let \(P_M \to M\) and \(P_A \to A\) be projective resolutions of \(M\) and \(A\) respectively. Then applying the Horseshoe Lemma, there is a projective resolution \(P_P \to P\) of \(P\) such that the following is a short exact sequence of chain complexes: \begin{align*} 0 \to P_M \to P_P \to P_A \to 0 ,\end{align*} where in fact in each degree \(n\) piece, this is induces a split exact sequence. Using that \(F\) is additive and additive functors preserve split exact sequences, the following is a SES for every \(n\): \begin{align*} 0 \to FP_M^n \to FP_P^n \to FP_A^n \to 0 ,\end{align*} which implies that there is a SES of chain complexes \begin{align*} 0 \to FP_M \to FP_P \to FP_A \to 0 .\end{align*} Thus there is an associated LES of derived functors:

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Using that \(P\) is \(F{\hbox{-}}\)acyclic, the middle terms \(L_i FP = 0\) for all \(i>0\), and thus this splits into a collection of SESs:

\begin{align*} 0 \to L_2 FA &\xrightarrow{{{\partial}}_2} L_1 FM \to 0\\ 0 \to L_3 FA &\xrightarrow{{{\partial}}_3} L_2 FM \to 0\\ &\vdots\\ 0 \to L_i FA &\xrightarrow{{{\partial}}_3} L_{i-1} FM \to 0 .\end{align*} This makes every \({{\partial}}_i\) for \(i\geq 2\) an isomorphism.

\begin{align*} L_iFA \cong \ker( FM \to FP) .\end{align*}

Using the same argument as above, consider the lower order terms of the associated LES:

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Noting that \(L_1FP = 0\) by \(F{\hbox{-}}\)acyclicity, the highlighted portion forms a four term exact sequence. We can form another exact sequence and compare the two:

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That the map indicated by the dotted line exists and is an isomorphism holds in any abelian category, using that fact that whenever \(0\to A \to B \xrightarrow{f} C\to 0\) is a SES we have \(A \cong \ker f\).

If \(P\to A\) is an \(F{\hbox{-}}\)acyclic resolution of \(A\), then there is an isomorphism \begin{align*} L_iFA \cong H_i(FP) .\end{align*}

\todo[inline]{Extend this to the exact sequence with $m$ terms, and show that the last conclusion holds.}

Show that the following are equivalent:

  • \(A\) is a projective \(R{\hbox{-}}\)module.

  • \(\mathop{\mathrm{Hom}}_R(A, {-})\) is an exact functor.

  • \(\operatorname{Ext} _R^{i \geq 1}(A, B) = 0\) and for all \(B\), i.e. \(A\) is \(\mathop{\mathrm{Hom}}_R({-}, B){\hbox{-}}\)acyclic for all \(B\).

  • \(\operatorname{Ext} _R^1(A, B)\) vanishes for all \(B\).

We’ll show

  • \(a \iff b\)
  • \(b\implies c\)
  • \(c\iff d\):
  • \(d \implies b\)

Let \(\xi\) be the following SES: \begin{align*} \xi: 0 \to M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3 \to 0 \end{align*} and define the functor \(F({-}) \coloneqq\mathop{\mathrm{Hom}}_R(A, {-})\). This is a covariant left-exact functor, and so applying it to the above sequence yields

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\(\implies\):

For \(F\) to be exact, it suffices to show it is right-exact, i.e. that \(F(g)\) is surjective. This amounts to asking that every \(\phi \in FM_3 \coloneqq\mathop{\mathrm{Hom}}_R(A, M_3)\) lifts to a preimage \(\tilde \phi \in FM_2 \coloneqq\mathop{\mathrm{Hom}}_R(A, M_2)\) satisfying \(F(g)(\tilde \phi) = \phi\). Unwinding definitions, this requires that \(g\circ \tilde \phi = \phi\), which is precisely the lift required for the universal property of projective objects:

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If \(A\) is projective, this lift always exists, so \(\mathop{\mathrm{Hom}}_R(A, {-})\) is an exact functor. Conversely, if \(\mathop{\mathrm{Hom}}_R(A, {-})\) is exact, this lift always exists, so \(A\) satisfies the universal property of a projective object.

Suppose \(F({-}) \coloneqq\mathop{\mathrm{Hom}}_R(A, {-})\) is exact, then since \(F\) is left-exact covariant it has right-derived functors \(\operatorname{Ext} _R^i(A, B) \coloneqq R^i F(B)\) which are computed in the following way

  • Taking an injective resolution of \begin{align*} B\to I \coloneqq(I_0 \xrightarrow{{{\partial}}_0} I_1 \xrightarrow{{{\partial}}_1} \cdots) .\end{align*}

  • Applying \(\mathop{\mathrm{Hom}}_R(A, {-})\) to get the complex \begin{align*} FI \coloneqq(0\to \mathop{\mathrm{Hom}}_R(A, I_0) \xrightarrow{F({{\partial}}_0)} \mathop{\mathrm{Hom}}_R(A, I_1) \xrightarrow{F({{\partial}}_1)} \cdots) .\end{align*}

  • Defining \begin{align*} R^iF(B) \coloneqq\ker F({{\partial}}_i) / \operatorname{im}F({{\partial}}_{i-1}) .\end{align*}

Note that in step (2), if \(\mathop{\mathrm{Hom}}_R(A, {-})\) is an exact functor, then since \(I\) is an acyclic complex, \(FI\) is again acyclic and so \(\ker F({{\partial}}_i) = \operatorname{im}F({{\partial}}_{i-1}) = 0\) for \(i\geq 1\). So \begin{align*} \operatorname{Ext} ^{\geq 1}_R(A, B) \coloneqq R^{\geq 1}F(B) = 0 .\end{align*}

\(\implies\): This direction is clear, since if \(\operatorname{Ext} ^i_R(A, B) = 0\) for all \(B\), then taking \(i=1\) is the statement of (d).

\

\(\impliedby\): This follows from the dimension-shifting isomorphism in a previous exercise. Let \(F({-}) \coloneqq\mathop{\mathrm{Hom}}_R(A, {-})\) and suppose \(\operatorname{Ext} ^1_R(A, B) \coloneqq L_1F(B) = 0\) for all \(B\). Let \(B'\) be arbitrary, it then suffices to show that \(\operatorname{Ext} ^i(A, B') \coloneqq L_i(B') = 0\) for all \(i> 1\), since we can take \(B'\) as one such \(B\) in the assumption for the \(i=1\) case.

The dimension shifting results states that if \(P_i\) are \(F{\hbox{-}}\)acyclic, then for every exact sequence \begin{align*} 0 \to M_m \to P_m \to \cdots \to P_0 \to B' \to 0 \end{align*} we obtain an isomorphism \begin{align*} L_iF(B') \cong L_{i-m-1}F(M_m) \iff L_{i}F(M_m) \cong L_{i+m+1}F(B') .\end{align*} So take any \(F{\hbox{-}}\)acyclic resolution of \(P\), say \begin{align*} B' \xrightarrow{{{\partial}}_{-1}} I_0 \xrightarrow{{{\partial}}_0} I_1 \xrightarrow{{{\partial}}_1} \cdots ,\end{align*} then consider truncating it at the \(m\)th stage: \begin{align*} 0 \to B' \xrightarrow{{{\partial}}_{i-1}} I_0 \to I_1 \to \cdots \xrightarrow{{{\partial}}_{m-1}} I_m \to M_m \coloneqq\operatorname{coker}{{\partial}}_{m-1} \to 0 \end{align*} By assumption, we have \(L_1F(M_m) = 0\) for every \(m\), and thus \begin{align*} 0 &= L_1F(B') \quad\text{by assumption} \\ 0 &= L_1F(M_0) \cong L_{2}F(B') \\ 0 &= L_1F(M_1) \cong L_{3}F(B') \\ 0 &= L_1F(M_2) \cong L_{4}F(B') \\ \vdots & \\ 0 &= L_1(M_m) \cong L_{m+2}(B') \quad \forall m\geq 0 .\end{align*} and so \(L_i(B') = 0\) for all \(i\geq 1\).

Take an arbitrary SES \begin{align*} \xi: 0\to B' \xrightarrow{f} B \xrightarrow{g} B'' \to 0 \end{align*} and consider applying the left-exact covariant functor \(F({-}) \coloneqq\mathop{\mathrm{Hom}}_R(A, {-})\) and taking the associated LES:

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By assumption, all of the higher $\operatorname{Ext} $ terms vanish, and in particular the red term \(\operatorname{Ext} ^1_R(A, B') = 0\). This implies that \(g^*\) is surjective, making the following sequence exact: \begin{align*} 0 \to {\mathop{\mathrm{Hom}}_R(A, B')} \xrightarrow{f^*} \mathop{\mathrm{Hom}}_R(A, B) \xrightarrow{g^*} {\mathop{\mathrm{Hom}}_R(A, B'')}\to 0 ,\end{align*} making \(\mathop{\mathrm{Hom}}_R(A, {-})\) an exact functor.

Show that \(\colim\) is left adjoint to \(\Delta\), and conclude that \(\colim\) is right-exact when when \(\mathsf{A}\) is abelian and \(\colim\) exists. Show that the pushout, i.e. \(\bullet \leftarrow \bullet \rightarrow\bullet\), is not an exact functor on \({\mathsf{Ab}}\).

Fixing some index category \(\mathsf{I}\) and a functor \(F: \mathsf{I} \to \mathsf{A}\), so \(F\in \mathsf{A}^{\mathsf{I}}\), write \(\widehat{A} \coloneqq\colim_{i \in I}F(i)\). We want to show that \(\adjunction{\colim}{\diagonal}{\mathsf{A}^{\mathsf{I}}}{\mathsf{A}}\) defines an adjoint pair, so that \(\colim\) is a left-adjoint and \(\diagonal\) is a right-adjoint. By definition, this is equivalent to showing the existence of natural bijections of sets \begin{align*} \tau_{FX}: \mathop{\mathrm{Hom}}_{\mathsf{A}}(\widehat{F}, X) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{\mathsf{A}^{\mathsf{I}} }(F, \Delta) \quad \forall X\in \mathsf{A}, F\in \mathsf{A}^{\mathsf{I}} .\end{align*}

We first note that the data of \(\widehat{F}\) is equivalent to the following universal property:

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That is, \(\widehat{F}\in {\operatorname{Ob}}(\mathsf{A})\) is an object equipped with structure maps \(\psi_i\) for every object \(F(i)\) in the image of \(F\) such that the solid triangle commutes, and for any object \(X\) with maps \(\phi_i: F(i)\to X, \phi_j:F(j)\to X\) making the outer triangle commute, there is a unique map \(\eta_{ij}: \widehat{F}\to X\) making the entire diagram commute. We can rewrite this condition in a more suggestive way:

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Applying the \(\diagonal\) functor, we can view this as a simpler universal property in \(\mathsf{A}^{\mathsf{I}}\), since the above data is precisely the data of a natural transformation:

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That is, the functor \(\Delta F\) is equipped with structure maps \(\Phi: F\to \Delta{F}\) which assemble into a natural transformation (i.e. a morphism in \(\mathsf{A}^{\mathsf{I}}\)) such that any other natural transformation from \(F\) to a diagonal object \(\Delta\) produces a unique natural transformation \(\eta: \Delta\to \Delta\). This provides exactly the data needed to specify \(\tau\): \begin{align*} \tau_{FX}: \mathop{\mathrm{Hom}}_{\mathsf{A}}(\widehat{F}, X) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{\mathsf{A}^{\mathsf{I}} }(F, \Delta) \\ \qty{ \widehat{F} \xrightarrow{f} X }&\mapsto \qty{ F \xrightarrow{\Psi} \Delta{F} \xrightarrow{\Delta f)} \Delta } ,\end{align*} i.e. we take the image \(\Delta(f)\) and pre-compose with the structure morphism \(\Psi\).

This is a bijection of sets.

This is surjective by the universal property: any morphism \(F \xrightarrow{g} \Delta\) in \(\mathsf{A}^{\mathsf{I}}\) factors through \(\Delta{F}\), and so all such morphisms are of this form.

\todo[inline]{Todo: showing surjectivity or constructing inverse.}

This is a natural isomorphism, i.e. for all \(X \xrightarrow{f} Y \in\mathop{\mathrm{Mor}}(\mathsf{A})\) and all \(F \xrightarrow{\eta} G \in \mathop{\mathrm{Mor}}(\mathsf{A}^{\mathsf{I}})\), there is a commuting diagram

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\todo[inline]{Show that the diagrams above always commute.}

If \(\mathsf{A}\) is abelian and \(\mathsf{I}\) is an index category such that \(\colim_{i\in \mathsf{I}}F(i)\) exists for all \(F\in \mathsf{A}^{\mathsf{I}}\), then the functor \(\colim: \mathsf{A}^{\mathsf{I}} \to \mathsf{A}\) is right-exact.

A sketch of the proof proceeds by showing every right adjoint is left-exact:

  • Since \(\mathop{\mathrm{Hom}}(LA, {-})\) is left-exact, we can apply it to a SES \(0\to B'\to B\to B''\to 0\).

  • Applying the natural isomorphisms coming from the adjunction, this is isomorphic to a sequence involving terms \(\mathop{\mathrm{Hom}}({-}, RB)\).

  • This sequence is exact, so applying Yoneda yields an exact sequence \begin{align*} 0\to RB' \to RB \to RB'' ,\end{align*} making \(R\) left-exact.

Finally, if \(L\) is a left adjoint out of \(\mathsf{A}\), then \(L^{\operatorname{op}}\) is a right adjoint out of \(\mathsf{A}^{\operatorname{op}}\). Thus \(L^{\operatorname{op}}\) is left-exact by the above argument, making \(L\) right-exact.

Let \(\mathsf{I} \coloneqq\qty{ \bullet \leftarrow \bullet \rightarrow\bullet}\) and define the pushout as \(\colim: \mathsf{A}^\mathsf{I} \to \mathsf{A}\). Then taking \(\mathsf{A} \coloneqq{\mathsf{Ab}}\), the pushout does not define an exact functor \(\mathsf{A}^\mathsf{I}\to \mathsf{A}\).

We proceed by constructing a counterexample. Unwinding definitions, we first note that an exact sequence of objects in \(\mathsf{A}^\mathsf{I}\) corresponds precisely to an exact sequence of diagrams. For pushouts, writing \(X_i\) for the pushout of \(A_i \mapsfrom P_i \mapsto B_i\), this gives an exact sequence of diagrams. If pushout were exact, this would in turn correspond to an exact sequence of the pushout objects \(X_i\) shown on the right:

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If we let \(f_i: P_i\to B_i\) be arbitrary maps between abelian groups and push out along \(A_i = 0\), we recover to cokernels of the \(f_i\):

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However, the sequence of cokernels appearing on the right is not exact in general, since this precisely fits into the diagram and exact sequence shown in the snake lemma:

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Here we know that the map involved in the red terms \(\operatorname{coker}f_1 \to \operatorname{coker}f_2\) is not injective in general, provided the green term \(\ker f_3\neq 0\). Thus an exact sequence of diagrams does not necessarily yield an exact sequence of their pushouts.