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

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:

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

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

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:

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

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

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

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

Since ${\mathbb{Z}}$ is a free ${\mathbb{Z}}{\hbox{-}}$module with generator 1, $f$ is entirely determined by $f(1)$, and
$f$ descends to a map $\tilde f: {\mathbb{Z}}/n \to {\mathbb{Q}}/{\mathbb{Z}}$ if and only if $f(n) \in {\mathbb{Z}}$, 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 $\mathop{\mathrm{Hom}}({\mathbb{Z}}/n, {\mathbb{Q}}/{\mathbb{Z}})$ as all of those morphism ${\mathbb{Z}}\to {\mathbb{Q}}/{\mathbb{Z}}$ which descend to make the following diagram commute.

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 $[0, 1) \cap{\mathbb{Q}}\subseteq {\mathbb{R}}$, 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 ${\mathbb{Z}}$) yields $n$ distinct morphisms $f_i: {\mathbb{Z}}\to {\mathbb{Q}}/{\mathbb{Z}}$ that descend to $\tilde f_i: {\mathbb{Z}}/n \to {\mathbb{Q}}/{\mathbb{Z}}$. We can define a map \begin{align*} \Psi: {\mathbb{Z}}&\to \mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/n, {\mathbb{Q}}/{\mathbb{Z}}) \\ i &\mapsto f_i ,\end{align*} and using the fact that if $i = i' \operatorname{mod}n$, write $i' = i + kn$ for some $k\in {\mathbb{Z}}$, 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 $k\in {\mathbb{Z}}$, so by the first isomorphism theorem $\Psi$ descends to an isomorphism \begin{align*} \tilde \Psi: {\mathbb{Z}}/n \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/n, {\mathbb{Q}}/{\mathbb{Z}}) .\end{align*}

${\mathbb{Q}}/{\mathbb{Z}}$ is an injective object in ${\mathbb{Z}}{\hbox{-}}$modules.

By the previous exercise, it suffices to show that ${\mathbb{Q}}/{\mathbb{Z}}$ 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 $n\in {\mathbb{Z}}$, 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 ${\mathbb{Q}}$ is divisible is a straightforward check: let $n\in {\mathbb{Z}}$ and $a\in {\mathbb{Q}}$, we then want a $b\in {\mathbb{Q}}$ such that $a = nb$, and $b \coloneqq{a\over n} \in {\mathbb{Q}}$ works. Since ${\mathbb{Q}}$ is an abelian group, ${\mathbb{Z}}$ is automatically normal, and the result follows.

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

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

$M \coloneqq{\mathbb{Z}}/m$,
$A \coloneqq{\mathbb{Z}}$,
$I \coloneqq n{\mathbb{Z}}$,

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

We can now use the map \begin{align*} \phi: {\mathbb{Z}}&\to {\mathbb{Z}}/m \otimes_{\mathbb{Z}}{\mathbb{Z}}/n \\ x & \mapsto x(1 \otimes 1) \end{align*} and compute \begin{align*} \ker \phi &= \left\{{x\in {\mathbb{Z}}{~\mathrel{\Big\vert}~}x(1\otimes 1) = 0}\right\}\\ &= \left\{{x\in {\mathbb{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 $F({-}) \coloneqq\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/p, {-})$, then noting that it is left-exact, it has right-derived functors.
Find an injective resolution $I$ of ${\mathbb{Z}}/q$.
Write $F(I)$ as a new (not necessarily exact) chain complex.
Compute $\operatorname{Ext} _{\mathbb{Z}}^i({\mathbb{Z}}/p, {\mathbb{Z}}/q) \coloneqq R^i F({\mathbb{Z}}/q) \coloneqq H^i(F({\mathbb{Z}}/q))$.

We can first take the following injective resolution:

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 $F({-})$:

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_{{\mathbb{Z}}}({\mathbb{Z}}/p, {\mathbb{Z}}/q) \coloneqq R^1 F({\mathbb{Z}}/q) \coloneqq{\ker {{\partial}}^1 \over \operatorname{im}{{\partial}}^0} = {{\mathbb{Z}}/p \over q \qty{{\mathbb{Z}}/p}} \cong {\mathbb{Z}}/ d{\mathbb{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, {\mathbb{Q}}/{\mathbb{Z}}) } {\mathbb{Q}}/{\mathbb{Z}}\\ a &\mapsto \mathbf{a} \text{ where } \mathbf{a}(f) \coloneqq f(a) \in {\mathbb{Q}}/{\mathbb{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 ${\mathbb{Q}}/{\mathbb{Z}}$ obtained by evaluating $f(a)$.

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

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

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

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

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:

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:

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 $f$:

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:

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:

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:

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

$\implies$:

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:

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:

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:

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:

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:

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

\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:

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 $f_i$:

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:

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.

Homological Algebra Problem Sets

Sunday, February 28th