--- title: "Homological Algebra Problem Sets" subtitle: "Problem Set 3" author: - name: D. Zack Garza affiliation: University of Georgia email: dzackgarza@gmail.com date: Spring 2021 order: 3 --- # Wednesday, February 17 :::{.problem title="Prove Corollary 2.3.2"} For $R$ a PID, show that an \(R\dash\)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 $b\in A$. ::: :::{.solution} 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 \(R\dash\)module morphism of the following form lifts, where we regard the ideal $J$ and the ring $R$ as \(R\dash\)modules: \begin{tikzcd} 0 && J && R \\ \\ && A \arrow[from=1-1, to=1-3] \arrow["f", from=1-3, to=3-3] \arrow["\iota", hook, from=1-3, to=1-5] \arrow["{\exists \tilde f}"{description}, dashed, from=1-5, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiSiJdLFs0LDAsIlIiXSxbMiwyLCJBIl0sWzAsMV0sWzEsMywiZiJdLFsxLDIsIlxcaW90YSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMywiXFxleGlzdHMgXFx0aWxkZSBmIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d) Since $R$ is a PID, we have $J = jR$ for some $j\in R$, so it suffices to produce lifts of the following form: \begin{tikzcd} 0 && jR && R \\ \\ && A \arrow[from=1-1, to=1-3] \arrow["f", from=1-3, to=3-3] \arrow["\iota", hook, from=1-3, to=1-5] \arrow["{\exists \tilde f}"{description}, dashed, from=1-5, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiSiJdLFs0LDAsIlIiXSxbMiwyLCJBIl0sWzAsMV0sWzEsMywiZiJdLFsxLDIsIlxcaW90YSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMywiXFxleGlzdHMgXFx0aWxkZSBmIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d) Consider $f(j)\in A$. Since $A$ is divisible, we have $A = jA$, so we can write $f(j) = j \vector a'$ for some $\vector a' \in A$. Using $R\dash$linearity and the fact that $j$ is left-cancelable, we have \[ jf(1_R) = f(j) = j\vector a' \implies f(1_R) = \vector a' .\] Thus we can set \[ \tilde f: R &\to A\\ 1_R &\mapsto \vector a' ,\] and extending $R\dash$linearly yields a well-defined \(R\dash\)module morphism. Moreover, the diagram commutes by construction, since $\iota(1_R) = 1_R$. --- $\impliedby$: Suppose $A\in\rmod$ is injective, where by Baer's criterion we equivalently have a lift of the following form for every $J\normal R$: \begin{tikzcd} 0 && J && R \\ \\ && A \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=3-3] \arrow[dashed, from=1-5, to=3-3] \arrow[hook, from=1-3, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiSiJdLFs0LDAsIlIiXSxbMiwyLCJNIl0sWzAsMV0sWzEsM10sWzIsMywiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzEsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=) 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 \( \vector a \in A \), there is a \( \vector a' \in A \) such that \( \vector a = j \vector a' \). Fixing $\vector{a}\in A$, define a map $f_a: J\to A$ in the following way: for $x\in J$, use the fact that \( \gens{ j }\da jR \) to first write $x = jr$ for some $r\in R$, and then set $f_a(x) = f_a(jr) \da r \vector{a}$. To summarize, we have \[ f_a: J = jR &\to R \\ x = jr &\mapsto r\vector{a} .\] By injectivity, we can take the inclusion $jR\injects R$ and get a lift: \begin{tikzcd} 0 && jR && R \\ \\ && A \arrow[from=1-1, to=1-3] \arrow["{f_a}", from=1-3, to=3-3] \arrow["\iota", hook, from=1-3, to=1-5] \arrow["{\exists \tilde f_a}"{description}, dashed, from=1-5, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwialIiXSxbNCwwLCJSIl0sWzIsMiwiQSJdLFswLDFdLFsxLDMsImZfYSJdLFsxLDIsIlxcaW90YSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMywiXFxleGlzdHMgXFx0aWxkZSBmX2EiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) We can now use the fact that \[ r \vector{a} &= f_a(jr) \\ &= \tilde f_a(\iota(jr)) \\ &= \tilde f_a(jr) \\ &= jr \tilde f_a(1_R) && \text{using $R\dash$linearity and }j,r\in R \\ &= rj \tilde f_a(1_R) && \text{since $R$ is commutative} \\ \implies \vector{a} &= j\tilde f_a(1_R) \in j A ,\] where in the last step we have canceled an $r$ on the left. So in the definition of divisibility, we can take \[ \vector a' \da \tilde f_a(1_R) ,\] and letting \( \vector a \) range over all elements of $A$ yields the desired result. ::: :::{.problem title="Calculating Ext Groups"} Calculate $\Ext_\ZZ^i(\ZZ/p, \ZZ/q)$ for distinct primes $p, q$. ::: The following are several claims that are later used in the actual solution: :::{.claim title="1"} For any $m\in \ZZ$, \[ \Hom_\ZZ(\ZZ/n, \QQ/\ZZ) \cong \ZZ/n .\] ::: :::{.proof title="?"} Note that there is an injection \[ 1 \to \Hom_\ZZ(\ZZ/n, \QQ/\ZZ) \injects \Hom_\ZZ(\ZZ, \QQ/\ZZ) ,\] which follows from the fact that there is a SES \[ 1 \to \ZZ \mapsvia{x\mapsto nx} \ZZ \mapsvia{\pi_n} \ZZ/n \to 1 \] where $\pi_m$ is the canonical quotient morphism, and applying the left-exact contravariant functor $\Hom_\ZZ(\ZZ/n, \QQ/\ZZ)$ yields the first exact sequence above. We use this to identify the former as a submodule of the latter, and note that for any \(\ZZ\dash\)module morphism $\ZZ \mapsvia{f} \QQ/\ZZ$, 1. Since $\ZZ$ is a free \(\ZZ\dash\)module with generator 1, $f$ is entirely determined by $f(1)$, and 2. $f$ descends to a map $\tilde f: \ZZ/n \to \QQ/\ZZ$ if and only if $f(n) \in \ZZ$, i.e. $f(n) = [0]$ is in the equivalence class of zero in the quotient, and so \[ [1] = [0] = f(n) = nf(1) .\] Using this injection, we can identify the submodule $\Hom(\ZZ/n, \QQ/\ZZ)$ as all of those morphism $\ZZ\to \QQ/\ZZ$ which descend to make the following diagram commute. \begin{tikzcd} \ZZ && {\QQ/\ZZ} \\ \\ {\ZZ/n} \arrow["{\exists \tilde f}"', color={rgb,255:red,92;green,92;blue,214}, dashed, from=3-1, to=1-3] \arrow["{\pi_m}"', from=1-1, to=3-1] \arrow["f", from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXFpaIl0sWzIsMCwiXFxRUS9cXFpaIl0sWzAsMiwiXFxaWi9tIl0sWzIsMSwiXFxleGlzdHMgXFx0aWxkZSBmIiwyLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMjQwLDYwLDYwLDFdXSxbMCwyLCJcXHBpX20iLDJdLFswLDEsImYiXV0=) 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 $[0, 1) \intersect \QQ \subseteq \RR$, where we want to find all $q \da f(1) \in [0, 1)$ such that $nq = 1$. A complete list of $n$ such representatives is given by \[ q \in \ts{ 0, {1\over n}, {2\over n}, \cdots, {n-1 \over n} } .\] Setting $f_i(1) \da \left[ {i\over n}\right]$ (where we take the equivalence class mod $\ZZ$) yields $n$ distinct morphisms $f_i: \ZZ\to \QQ/\ZZ$ that descend to $\tilde f_i: \ZZ/n \to \QQ/\ZZ$. We can define a map \[ \Psi: \ZZ &\to \Hom_\ZZ(\ZZ/n, \QQ/\ZZ) \\ i &\mapsto f_i ,\] and using the fact that if $i = i' \mod n$, write $i' = i + kn$ for some $k\in \ZZ$, then \[ f_{i'}(1) = f_{i + kn}(1) = \left[ {i + kn \over n}\right] = \left[{i\over n} + k\right] = \left[ {i\over n} \right] = f_{i}(1) ,\] since $k\in \ZZ$, so by the first isomorphism theorem $\Psi$ descends to an isomorphism \[ \tilde \Psi: \ZZ/n \mapsvia{\sim} \Hom_\ZZ(\ZZ/n, \QQ/\ZZ) .\] ::: :::{.claim title="2"} $\QQ/\ZZ$ is an injective object in \(\ZZ\dash\)modules. ::: :::{.proof title="?"} By the previous exercise, it suffices to show that $\QQ/\ZZ$ is divisible. More generally, if any group $G$ is divisible and $N\normal G$ is a normal subgroup, then $G/N$ will be divisible. This follows from the fact that if $\bar{a}, \bar{b} \in G/N$ and $n\in \ZZ$, we can write $\bar a = a + N$ and $\bar b = b + N$ for some coset representatives, use divisibility to write $a = nb$, and then compute \[ \bar a = a + N = (nb) + N \da n(b + N) = n \bar{b} .\] That $\QQ$ is divisible is a straightforward check: let $n\in \ZZ$ and $a\in \QQ$, we then want a $b\in \QQ$ such that $a = nb$, and $b \da {a\over n} \in \QQ$ works. Since $\QQ$ is an abelian group, $\ZZ$ is automatically normal, and the result follows. ::: :::{.claim} \[ {\ZZ/n \over m(\ZZ/n)} \cong \ZZ/d && d \da \gcd(\ZZ/m, \ZZ/n) .\] ::: :::{.proof title="?"} Using \[ M\tensor_R {A\over I} \cong {M\over IM} \in \rmod ,\] and taking - $M \da \ZZ/m$, - $A \da \ZZ$, - $I \da n\ZZ$, we have \[ \ZZ/m \tensor_\ZZ \ZZ/n \cong {\ZZ/m \over n\qty{\ZZ/m} } && \in \mods{\ZZ} .\] We can now use the map \[ \phi: \ZZ &\to \ZZ/m \tensor_\ZZ \ZZ/n \\ x & \mapsto x(1 \tensor 1) \] and compute \[ \ker \phi &= \ts{x\in \ZZ \st x(1\tensor 1) = 0}\\ &= \ts{x\in \ZZ \st n\divides x \text{ or } m \divides x}\\ &= \gens{ n, m } \\ &= \gens{ \gcd(n, m) } && \text{by Bezout's theorem} \\ &\da \gens{ d } .\] Now applying the first isomorphism theorem yields the result. ::: :::{.solution} We'll follow the procedure outlined in Weibel: - Define the contravariant functor $F(\wait) \da \Hom_\ZZ(\ZZ/p, \wait)$, then noting that it is left-exact, it has right-derived functors. - Find an injective resolution $I$ of $\ZZ/q$. - Write $F(I)$ as a new (not necessarily exact) chain complex. - Compute $\Ext_\ZZ^i(\ZZ/p, \ZZ/q) \da R^i F(\ZZ/q) \da H^i(F(\ZZ/q))$. We can first take the following injective resolution: \begin{tikzcd} {1 } && {\ZZ/q} && {\QQ/\ZZ} && {\QQ/\ZZ} && 1 \\ && {[1]_q} && {\left[1\over q \right]} \\ \\ &&&& {[x]} && {[qx]} \arrow[from=1-1, to=1-3] \arrow["{d^{-1}}", from=1-3, to=1-5] \arrow["{d^0}", from=1-5, to=1-7] \arrow["{d^1}", from=1-7, to=1-9] \arrow[from=2-3, to=2-5] \arrow[from=4-5, to=4-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMCwwLCIxICJdLFsyLDAsIlxcWlovcSJdLFs0LDAsIlxcUVEvXFxaWiJdLFs2LDAsIlxcUVEvXFxaWiJdLFs4LDAsIjEiXSxbMiwxLCJbMV1fcSJdLFs0LDEsIlxcbGVmdFsxXFxvdmVyIG4gXFxyaWdodF0iXSxbNCwzLCJbeF0iXSxbNiwzLCJbbnhdIl0sWzAsMV0sWzEsMiwiZF57LTF9Il0sWzIsMywiZF4wIl0sWzMsNCwiZF4xIl0sWzUsNl0sWzcsOF1d) 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 $F(\wait)$: \begin{tikzcd} 1 && {I^0\da \QQ/\ZZ} && {I^1 \da \QQ/\ZZ} && 1 \\ \\ \\ 1 && {F(I^0) \da \Hom_\ZZ(\ZZ/p, \QQ/\ZZ)} && {F(I^1) \da \Hom_\ZZ(\ZZ/p, \QQ/\ZZ)} && 1 \\ \\ 1 && {\ZZ/p} && {\ZZ/p} && 1 \arrow[from=1-1, to=1-3] \arrow[""{name=0, anchor=center, inner sep=0}, "{d^0}", from=1-3, to=1-5] \arrow["{d^1}", from=1-5, to=1-7] \arrow[from=4-1, to=4-3] \arrow[""{name=1, anchor=center, inner sep=0}, "{\bd^0 \da F(d^0)}", from=4-3, to=4-5] \arrow["{\bd^1 \da F(d^1)}", from=4-5, to=4-7] \arrow["{\tilde \bd^1}", from=6-5, to=6-7] \arrow["{\tilde \bd^0}", from=6-3, to=6-5] \arrow[from=6-1, to=6-3] \arrow["{\Psi \cong}"{description}, from=6-3, to=4-3] \arrow["{\Psi \cong}"{description}, from=6-5, to=4-5] \arrow[equal, from=6-1, to=4-1] \arrow[equal, from=6-7, to=4-7] \arrow["{F(\wait)}", shorten <=13pt, shorten >=13pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Here we immediately simplify by applying the isomorphism from the earlier claim. Noting that $d^0(x) \da qx$ was multiplication by $q$, we have $\bd^0(f) = d^0 \circ f$ is post-composition by the multiplication by $q$ map, and $\tilde \bd^0$ similarly becomes multiplication by $q$. We now take homology: \[ \Ext^1_{\ZZ}(\ZZ/p, \ZZ/q) \da R^1 F(\ZZ/q) \da {\ker \bd^1 \over \im \bd^0} = {\ZZ/p \over q \qty{\ZZ/p}} \cong \ZZ/ d\ZZ \cong 1 ,\] where $d \da \gcd(p, q) = 1$ if $p, q$ are coprime. ::: :::{.problem title="Weibel 2.3.2"} Let $A\in \Ab$, and show that the following map is injective: \[ \eps_A : A &\to I(A) \da \prod_{f\in \Hom_\Ab(A, \QQ/\ZZ) } \QQ/\ZZ \\ a &\mapsto \vector a \text{ where } \vector{a}(f) \da f(a) \in \QQ/\ZZ ,\] i.e. when looking at the image $\eps_A(a)$ in the product, the component indexed by $f$ is an element of $\QQ/\ZZ$ obtained by evaluating $f(a)$. *Hint: if $a\in A$, find a map $f: a \ZZ\to \QQ/\ZZ$ with $f(a) \neq 0$ and extend this to a map $f': A\to \QQ/\ZZ$.* ::: :::{.solution} By contrapositive, we'll suppose $a\neq 0$ and show $\eps_A(a) \neq 0$. Following the hint, we first consider the cyclic subgroup $a\ZZ = \ts{an \st n\in \ZZ}$ and define a map \[ f_a: a\ZZ &\to \ZZ \\ an & \mapsto n .\] We now pick $\ell > 1 \in \ZZ$ to be any integer, and define a composition $f: a\ZZ \to \QQ/\ZZ$: \begin{tikzcd} a\ZZ && \ZZ && \QQ && \QQ && {\QQ/\ZZ} \\ an && n && n && {{n \over \ell}} && {\left[ n\over \ell \right]} \\ a &&&&&&&& {\left[1\over \ell\right]} \arrow["{f_a}", from=1-1, to=1-3] \arrow["\iota", hook, from=1-3, to=1-5] \arrow["{x\mapsto {x\over \ell}}", from=1-5, to=1-7] \arrow["\pi", from=1-7, to=1-9] \arrow[maps to, from=2-1, to=2-3] \arrow[maps to, from=2-3, to=2-5] \arrow[maps to, from=2-5, to=2-7] \arrow[maps to, from=2-7, to=2-9] \arrow[color={rgb,255:red,92;green,214;blue,92}, curve={height=-30pt}, dashed, from=1-1, to=1-9, "{f}"] \arrow[color={rgb,255:red,92;green,214;blue,92}, dashed, from=3-1, to=3-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) By choice of $\ell$, this map satisfies $f(a) = [1/\ell] \neq 0$, so the map is nonzero. Since $\QQ/\ZZ$ is injective, the universal property provides a lift $\tilde f$: \begin{tikzcd} A \\ \\ a\ZZ && {\QQ/\ZZ} \arrow["f", from=3-1, to=3-3] \arrow[hook, from=3-1, to=1-1] \arrow["{\exists \tilde f}", dashed, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwyLCJhXFxaWiJdLFswLDAsIkEiXSxbMiwyLCJcXFFRL1xcWloiXSxbMCwyLCJmIl0sWzAsMSwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwyLCJcXGV4aXN0cyBcXHRpbGRlIGYiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) Since $\tilde f$ lifts $f$, it is also nonzero. But now we can check that \[ \eps_A(a)(f) \da f(a) \neq 0 ,\] so the $f$ component of the image of $a$ is nonzero and thus $\vector a \da \eps_A(a) \neq 0$ in the product. ::: # Sunday, February 28th :::{.problem title="Weibel 2.4.2"} If $U: \cat{B} \to \cat{C}$ is right-exact functor, show that \[ U(L_i F) \cong L_i(UF) .\] ::: :::{.solution} 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: \begin{tikzcd} \cdots && {P_2} && {P_1} && {P_0} && X && 0 \\ \\ \cdots && {FP_2} && {FP_1} && {FP_0} && 0 \arrow["\eps", from=1-7, to=1-9] \arrow["0", from=1-9, to=1-11] \arrow[""{name=0, anchor=center, inner sep=0}, "{\bd_1}", from=1-5, to=1-7] \arrow["{\bd_2}", from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \arrow[from=3-1, to=3-3] \arrow["{F(\bd_2)}"', from=3-3, to=3-5] \arrow[""{name=1, anchor=center, inner sep=0}, "{F(\bd_1)}"', from=3-5, to=3-7] \arrow["0"', from=3-7, to=3-9] \arrow["{F(\wait)}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTEsWzIsMCwiUF8yIl0sWzQsMCwiUF8xIl0sWzYsMCwiUF8wIl0sWzgsMCwiWCJdLFsxMCwwLCIwIl0sWzAsMCwiXFxjZG90cyJdLFs2LDIsIkZQXzAiXSxbNCwyLCJGUF8xIl0sWzIsMiwiRlBfMiJdLFswLDIsIlxcY2RvdHMiXSxbOCwyLCIwIl0sWzIsMywiXFxlcHMiXSxbMyw0LCIwIl0sWzEsMiwiXFxiZF8xIl0sWzAsMSwiXFxiZF8yIl0sWzUsMF0sWzksOF0sWzgsNywiRihcXGJkXzIpIl0sWzcsNiwiRihcXGJkXzEpIl0sWzYsMTAsIjAiXSxbMTMsMTgsIkYoXFx3YWl0KSIsMCx7InNob3J0ZW4iOnsic291cmNlIjoyMCwidGFyZ2V0IjoyMH19XV0=) We now have by definition \[ L_iF(X) \da { \ker F(\bd_i) \over \im F(\bd_{i+1})} \implies U\qty{ L_iF(X)} \da U\qty{ \ker F(\bd_i) \over \im F(\bd_{i+1})} .\] For the right-hand side, we can take the same projective resolution $P \to X$, and apply a similar process: \begin{tikzcd} \cdots && {P_2} && {P_1} && {P_0} && X && 0 \\ \\ \cdots && {UFP_2} && {UFP_1} && {UFP_0} && 0 \arrow["\eps", from=1-7, to=1-9] \arrow["0", from=1-9, to=1-11] \arrow[""{name=0, anchor=center, inner sep=0}, "{\bd_1}", from=1-5, to=1-7] \arrow["{\bd_2}", from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \arrow[from=3-1, to=3-3] \arrow["{(UF)(\bd_2)}"', from=3-3, to=3-5] \arrow[""{name=1, anchor=center, inner sep=0}, "{(UF)(\bd_1)}"', from=3-5, to=3-7] \arrow["0"', from=3-7, to=3-9] \arrow["{(U\circ F)(\wait)}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTEsWzIsMCwiUF8yIl0sWzQsMCwiUF8xIl0sWzYsMCwiUF8wIl0sWzgsMCwiWCJdLFsxMCwwLCIwIl0sWzAsMCwiXFxjZG90cyJdLFs2LDIsIlVGUF8wIl0sWzQsMiwiVUZQXzEiXSxbMiwyLCJVRlBfMiJdLFswLDIsIlxcY2RvdHMiXSxbOCwyLCIwIl0sWzIsMywiXFxlcHMiXSxbMyw0LCIwIl0sWzEsMiwiXFxiZF8xIl0sWzAsMSwiXFxiZF8yIl0sWzUsMF0sWzksOF0sWzgsNywiKFVGKShcXGJkXzIpIiwyXSxbNyw2LCIoVUYpKFxcYmRfMSkiLDJdLFs2LDEwLCIwIiwyXSxbMTMsMTgsIihVXFxjaXJjIEYpKFxcd2FpdCkiLDAseyJzaG9ydGVuIjp7InNvdXJjZSI6MjAsInRhcmdldCI6MjB9fV1d) Again, by definition, \[ (L_i(UF))(X) \da { \ker (UF)(\bd_i) \over \im (UF)(\bd_{i+1})} ,\] and thus it suffices to show that there is an isomorphism \[ U\qty{ \ker F(\bd_i) \over \im F(\bd_{i+1})} \mapsvia{\sim} { \ker (UF)(\bd_i) \over \im (UF)(\bd_{i+1})} .\] 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 $f$: \begin{tikzcd}[column sep=1em] 0 && {\im F(\bd_{i+1})} && {\ker F(\bd_{i})} && {{\ker F(\bd_{i}) \over \im F(\bd_{i+1})}} && 0 \\ \\ 0 && {U\qty{\im F(\bd_{i+1})}} && {U\qty{\ker F(\bd_{i})}} && \textcolor{rgb,255:red,92;green,92;blue,214}{U\qty{\ker F(\bd_{i}) \over \im F(\bd_{i+1})}} && 0 \\ \\ 0 && {U\qty{\im F(\bd_{i+1})}} && {U\qty{\ker F(\bd_{i})}} && \textcolor{rgb,255:red,92;green,92;blue,214}{{U(\ker F(\bd_{i})) \over U(\im F(\bd_{i+1}))}} && 0 \arrow[from=1-1, to=1-3] \arrow["{\iota_i}", from=1-3, to=1-5] \arrow["{\pi_i}", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow["{U(\iota_i)}", from=3-3, to=3-5] \arrow["{U(\pi_i)}", from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[equal, from=3-5, to=5-5] \arrow[equal, from=3-3, to=5-3] \arrow["f", color={rgb,255:red,92;green,214;blue,92}, from=3-7, to=5-7] \arrow[from=5-1, to=5-3] \arrow["{U(\iota_i)}", from=5-3, to=5-5] \arrow["{\tilde \pi_i}", from=5-5, to=5-7] \arrow[from=5-7, to=5-9] \arrow["U", color={rgb,255:red,214;green,92;blue,92}, shorten <=6pt, shorten >=6pt, Rightarrow, from=1-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Here $\tilde \pi_i$ is the natural quotient map, whose image is $\coker U(\iota_i)$. Finally, the map $f$ exists an any abelian category, using that whenever $0\to A \mapsvia{g_1} B \mapsvia{g_2} C\to 0$ is exact, there is an isomorphism $C \mapsvia{\sim} B/\im(g_1)$. ::: :::{.problem title="Weibel 2.4.3"} \envlist - If $0\to M \to P \to A \to 0$ is exact with $P$ projective or $F\dash$acyclic, show that \[ L_i F(A) \cong L_{i-1}FM && i\geq 2 .\] - Show that if \[ 0 \to M_m \to P_m \to P_{m-1} \to \cdots \to P_0 \to A \to 0 \] is exact with $P_i$ projective or $F\dash$acyclic, then \[ L_iF(A) \cong L_{i-m-1}F(M_m) && i\geq m+2 .\] - 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 $F\dash$acyclic resolution of $A$, then $L_i F(A) = H_i(F(P))$. ::: :::{.solution} \envlist :::{.claim} \[ L_i F(A) \cong L_{i-1}FM && i\geq 2 .\] ::: :::{.proof title="of claim"} 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: \[ 0 \to P_M \to P_P \to P_A \to 0 ,\] 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 $n$: \[ 0 \to FP_M^n \to FP_P^n \to FP_A^n \to 0 ,\] which implies that there is a SES of chain complexes \[ 0 \to FP_M \to FP_P \to FP_A \to 0 .\] Thus there is an associated LES of derived functors: \begin{tikzcd} &&&& \cdots \\ \\ {L_iFM} && \textcolor{rgb,255:red,214;green,92;blue,92}{L_iFP} && {L_iFA} \\ \\ {L_{i-1}FM} && \textcolor{rgb,255:red,214;green,92;blue,92}{L_{i-1}FP} && \cdots \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow["{\bd_i}", from=3-5, to=5-1, in=180, out=0] \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow["{\bd_{i+1}}", from=1-5, to=3-1, in=180, out=0] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwyLCJMX2lGTSJdLFsyLDIsIkxfaUZQIixbMCw2MCw2MCwxXV0sWzQsMiwiTF9pRkEiXSxbMCw0LCJMX3tpLTF9Rk0iXSxbMiw0LCJMX3tpLTF9RlAiLFswLDYwLDYwLDFdXSxbNCw0LCJcXGNkb3RzIl0sWzQsMCwiXFxjZG90cyJdLFswLDFdLFsxLDJdLFsyLDMsIlxcYmRfaSJdLFszLDRdLFs0LDVdLFs2LDAsIlxcYmRfe2krMX0iXV0=) Using that $P$ is $F\dash$acyclic, the middle terms $L_i FP = 0$ for all $i>0$, and thus this splits into a collection of SESs: \[ 0 \to L_2 FA &\mapsvia{\bd_2} L_1 FM \to 0\\ 0 \to L_3 FA &\mapsvia{\bd_3} L_2 FM \to 0\\ &\vdots\\ 0 \to L_i FA &\mapsvia{\bd_3} L_{i-1} FM \to 0 .\] This makes every $\bd_i$ for $i\geq 2$ an isomorphism. ::: :::{.claim} \[ L_iFA \cong \ker( FM \to FP) .\] ::: :::{.proof title="?"} Using the same argument as above, consider the lower order terms of the associated LES: \begin{tikzcd} &&&& \cdots \\ \\ {L_1FM} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_1FP = 0} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_1FA} \\ \\ \textcolor{rgb,255:red,92;green,214;blue,92}{L_{0}FM} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_{0}FP} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_0FA} & \textcolor{rgb,255:red,92;green,214;blue,92}{0} \arrow[from=3-1, to=3-3] \arrow[color={rgb,255:red,92;green,214;blue,92}, from=3-3, to=3-5] \arrow["{\bd_1}", color={rgb,255:red,92;green,214;blue,92}, from=3-5, to=5-1] \arrow[color={rgb,255:red,92;green,214;blue,92}, from=5-1, to=5-3] \arrow[color={rgb,255:red,92;green,214;blue,92}, from=5-3, to=5-5] \arrow[from=5-5, to=5-6] \arrow["{\bd_2}", from=1-5, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Noting that $L_1FP = 0$ by $F\dash$acyclicity, the highlighted portion forms a four term exact sequence. We can form another exact sequence and compare the two: \begin{tikzcd}[column sep=1em] &&&&&&&& {} \\ \\ 0 && \textcolor{rgb,255:red,92;green,214;blue,92}{L_1FA} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_{0}FM} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_{0}FP} && \textcolor{rgb,255:red,92;green,214;blue,92}{L_0FA} && \textcolor{rgb,255:red,92;green,214;blue,92}{0} \\ \\ 0 && {\ker(L_0 FM \to L_0FP)} && {L_{0}FM} && {L_{0}FP} && {L_0FA} && 0 \arrow[color={rgb,255:red,92;green,214;blue,92}, from=3-5, to=3-7] \arrow[color={rgb,255:red,92;green,214;blue,92}, from=3-7, to=3-9] \arrow["{\bd_1}", color={rgb,255:red,92;green,214;blue,92}, from=3-3, to=3-5] \arrow[from=3-1, to=3-3] \arrow[color={rgb,255:red,92;green,214;blue,92}, from=3-9, to=3-11] \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow[from=5-5, to=5-7] \arrow[from=5-7, to=5-9] \arrow[from=5-9, to=5-11] \arrow[equal, from=3-9, to=5-9] \arrow[equal, from=3-7, to=5-7] \arrow[equal, from=3-5, to=5-5] \arrow[dashed, from=3-3, to=5-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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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 $0\to A \to B \mapsvia{f} C\to 0$ is a SES we have $A \cong \ker f$. ::: :::{.claim} If $P\to A$ is an $F\dash$acyclic resolution of $A$, then there is an isomorphism \[ L_iFA \cong H_i(FP) .\] ::: :::{.proof title="?"} \todo[inline]{Extend this to the exact sequence with $m$ terms, and show that the last conclusion holds.} ::: ::: :::{.problem title="Weibel 2.5.2"} Show that the following are equivalent: a. $A$ is a projective \(R\dash\)module. b. $\Hom_R(A, \wait)$ is an exact functor. c. $\Ext_R^{i \geq 1}(A, B) = 0$ and for all $B$, i.e. $A$ is $\Hom_R(\wait, B)\dash$acyclic for all $B$. d. $\Ext_R^1(A, B)$ vanishes for all $B$. ::: :::{.solution} We'll show - $a \iff b$ - $b\implies c$ - $c\iff d$: - $d \implies b$ ::: :::{.proof title="$a\iff b$"} Let $\xi$ be the following SES: \[ \xi: 0 \to M_1 \mapsvia{f} M_2 \mapsvia{g} M_3 \to 0 \] and define the functor $F(\wait) \da \Hom_R(A, \wait)$. This is a covariant left-exact functor, and so applying it to the above sequence yields \begin{tikzcd} 0 && {M_1} && {M_2} && {M_3} && 0 \\ \\ 0 && {FM_1} && {FM_2} && {FM_3} && {} \arrow[from=1-1, to=1-3] \arrow["f", from=1-3, to=1-5] \arrow["g", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow["{F(f): \lambda \mapsto f\circ \lambda}"', from=3-3, to=3-5] \arrow["{F(g): \lambda \mapsto g\circ \lambda}"', from=3-5, to=3-7] \arrow["{F(\wait) \da \Hom_R(A, \wait)}"{description}, Rightarrow, from=1-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIk1fMSJdLFs0LDAsIk1fMiJdLFs2LDAsIk1fMyJdLFs4LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiRk1fMSJdLFs0LDIsIkZNXzIiXSxbNiwyLCJGTV8zIl0sWzgsMl0sWzAsMV0sWzEsMiwiZiJdLFsyLDMsImciXSxbMyw0XSxbNSw2XSxbNiw3LCJGKGYpOiBcXGxhbWJkYSBcXG1hcHN0byBmXFxjaXJjIFxcbGFtYmRhIiwyXSxbNyw4LCJGKGcpOiBcXGxhbWJkYSBcXG1hcHN0byBnXFxjaXJjIFxcbGFtYmRhIiwyXSxbMiw3LCJGKFxcd2FpdCkgXFxkYSBcXEhvbV9SKEEsIFxcd2FpdCkiLDEseyJsZXZlbCI6Mn1dXQ==) $\implies$: 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 $\phi \in FM_3 \da \Hom_R(A, M_3)$ lifts to a preimage $\tilde \phi \in FM_2 \da \Hom_R(A, M_2)$ 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: \begin{tikzcd} && A \\ \\ {M_2} && {M_3} && 0 \arrow["g", from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow["\phi", from=1-3, to=3-3] \arrow["{\exists \tilde \phi}"', color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-3, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJNXzIiXSxbMiwyLCJNXzMiXSxbNCwyLCIwIl0sWzIsMCwiQSJdLFswLDEsImciXSxbMSwyXSxbMywxLCJcXHBoaSJdLFszLDAsIlxcZXhpc3RzIFxcdGlsZGUgXFxwaGkiLDIseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19LFsyNDAsNjAsNjAsMV1dXQ==) If $A$ is projective, this lift always exists, so $\Hom_R(A, \wait)$ is an exact functor. Conversely, if $\Hom_R(A, \wait)$ is exact, this lift always exists, so $A$ satisfies the universal property of a projective object. ::: :::{.proof title="$b\implies c$"} Suppose $F(\wait) \da \Hom_R(A, \wait)$ is exact, then since $F$ is left-exact covariant it has right-derived functors $\Ext_R^i(A, B) \da R^i F(B)$ which are computed in the following way 1. Taking an *injective* resolution of \[ B\to I \da (I_0 \mapsvia{\bd_0} I_1 \mapsvia{\bd_1} \cdots) .\] 2. Applying $\Hom_R(A, \wait)$ to get the complex \[ FI \da (0\to \Hom_R(A, I_0) \mapsvia{F(\bd_0)} \Hom_R(A, I_1) \mapsvia{F(\bd_1)} \cdots) .\] 3. Defining \[ R^iF(B) \da \ker F(\bd_i) / \im F(\bd_{i-1}) .\] Note that in step (2), if $\Hom_R(A, \wait)$ is an exact functor, then since $I$ is an acyclic complex, $FI$ is again acyclic and so $\ker F(\bd_i) = \im F(\bd_{i-1}) = 0$ for $i\geq 1$. So \[ \Ext^{\geq 1}_R(A, B) \da R^{\geq 1}F(B) = 0 .\] ::: :::{.proof title="$c\iff d$"} $\implies$: This direction is clear, since if $\Ext^i_R(A, B) = 0$ 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 $F(\wait) \da \Hom_R(A, \wait)$ and suppose $\Ext^1_R(A, B) \da L_1F(B) = 0$ for all $B$. Let $B'$ be arbitrary, it then suffices to show that $\Ext^i(A, B') \da L_i(B') = 0$ 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 $F\dash$acyclic, then for every exact sequence \[ 0 \to M_m \to P_m \to \cdots \to P_0 \to B' \to 0 \] we obtain an isomorphism \[ L_iF(B') \cong L_{i-m-1}F(M_m) \iff L_{i}F(M_m) \cong L_{i+m+1}F(B') .\] So take any $F\dash$acyclic resolution of $P$, say \[ B' \mapsvia{\bd_{-1}} I_0 \mapsvia{\bd_0} I_1 \mapsvia{\bd_1} \cdots ,\] then consider truncating it at the $m$th stage: \[ 0 \to B' \mapsvia{\bd_{i-1}} I_0 \to I_1 \to \cdots \mapsvia{\bd_{m-1}} I_m \to M_m \da \coker \bd_{m-1} \to 0 \] By assumption, we have $L_1F(M_m) = 0$ for every $m$, and thus \[ 0 &= L_1F(B') \quad\text{by assumption} \\ 0 &= L_1F(M_0) \cong L_{2}F(B') \\ 0 &= L_1F(M_1) \cong L_{3}F(B') \\ 0 &= L_1F(M_2) \cong L_{4}F(B') \\ \vdots & \\ 0 &= L_1(M_m) \cong L_{m+2}(B') \quad \forall m\geq 0 .\] and so $L_i(B') = 0$ for all $i\geq 1$. ::: :::{.proof title="$d\implies b$"} Take an arbitrary SES \[ \xi: 0\to B' \mapsvia{f} B \mapsvia{g} B'' \to 0 \] and consider applying the left-exact covariant functor $F(\wait) \da \Hom_R(A, \wait)$ and taking the associated LES: \begin{tikzcd} 0 \\ {\Hom_R(A, B')} & \Hom_R(A, B) & {\Hom_R(A, B'')} \\ \\ \textcolor{rgb,255:red,214;green,92;blue,92}{\Ext^1_R(A, B')} & {\Ext^1_R(A, B)} & {\Ext^1_R(A, B'')} \\ \\ \cdots \arrow[from=1-1, to=2-1] \arrow["f^*", from=2-1, to=2-2] \arrow["g^*", from=2-2, to=2-3] \arrow["{\delta_0}", from=2-3, to=4-1, in=180, out=0] \arrow[from=4-1, to=4-2] \arrow[from=4-2, to=4-3] \arrow["{\delta_1}", from=4-3, to=6-1, in=180, out=0] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJCJyJdLFsxLDEsIkIiXSxbMiwxLCJCJyciXSxbMCwzLCJcXEV4dF4xX1IoQSwgQicpIixbMCw2MCw2MCwxXV0sWzEsMywiXFxFeHReMV9SKEEsIEIpIl0sWzIsMywiXFxFeHReMV9SKEEsIEInJykiXSxbMCwwLCIwIl0sWzAsNSwiXFxjZG90cyJdLFs2LDBdLFswLDEsImYiXSxbMSwyLCJnIl0sWzIsMywiXFxkZWx0YV8wIl0sWzMsNF0sWzQsNV0sWzUsNywiXFxkZWx0YV8xIl1d) By assumption, all of the higher $\Ext$ terms vanish, and in particular the red term $\Ext^1_R(A, B') = 0$. This implies that $g^*$ is surjective, making the following sequence exact: \[ 0 \to {\Hom_R(A, B')} \mapsvia{f^*} \Hom_R(A, B) \mapsvia{g^*} {\Hom_R(A, B'')}\to 0 ,\] making $\Hom_R(A, \wait)$ an exact functor. ::: :::{.problem title="Weibel 2.6.4"} Show that $\colim$ is left adjoint to $\Delta$, and conclude that $\colim$ is right-exact when when \( \cat{A} \) is abelian and $\colim$ exists. Show that the pushout, i.e. $\bullet \leftarrow \bullet \rightarrow\bullet$, is not an exact functor on $\Ab$. ::: :::{.solution} Fixing some index category $\cat{I}$ and a functor $F: \cat{I} \to \cat{A}$, so $F\in \cat{A}^{\cat{I}}$, write $\hat{A} \da \colim_{i \in I}F(i)$. We want to show that $\adjunction{\colim}{\diagonal}{\cat{A}^{\cat{I}}}{\cat{A}}$ 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 \[ \tau_{FX}: \Hom_{\cat{A}}(\hat{F}, X) \mapsvia{\sim} \Hom_{\cat{A}^{\cat{I}} }(F, \diagonal X) \quad \forall X\in \cat{A}, F\in \cat{A}^{\cat{I}} .\] We first note that the data of $\hat F$ is equivalent to the following universal property: \begin{tikzcd} {\cat{I}} &&& {\cat{A}} \\ \\ i &&& {F(i)} \\ &&&&&& {\hat{F}} && X \\ j &&& {F(j)} \arrow[""{name=0, anchor=center, inner sep=0}, "{\eps_{ij}}"', from=3-1, to=5-1] \arrow[""{name=1, anchor=center, inner sep=0}, "{F\eps_{ij}}"', from=3-4, to=5-4] \arrow["{\psi_i}", from=3-4, to=4-7] \arrow["{\psi_j}"', from=5-4, to=4-7] \arrow["{\phi_i}", curve={height=-12pt}, dotted, from=3-4, to=4-9] \arrow["{\phi_j}"', curve={height=12pt}, dotted, from=5-4, to=4-9] \arrow["{\exists ! \eta_{ij}}"{description, pos=0.4}, dashed, from=4-7, to=4-9] \arrow["{F(\wait)}"', shorten <=30pt, shorten >=30pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) That is, $\hat{F}\in \Ob(\cat{A})$ 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 $\eta_{ij}: \hat{F}\to X$ making the entire diagram commute. We can rewrite this condition in a more suggestive way: \begin{tikzcd} {\cat{I}} &&& {\cat{A}} \\ \\ i &&& {F(i)} &&& {\hat{F}} &&& X \\ \\ j &&& {F(j)} &&& {\hat{F}} &&& X \arrow[""{name=0, anchor=center, inner sep=0}, "{\eps_{ij}}"', from=3-1, to=5-1] \arrow[""{name=1, anchor=center, inner sep=0}, "{F\eps_{ij}}"', from=3-4, to=5-4] \arrow["{\psi_i}", from=3-4, to=3-7] \arrow["{\psi_j}"', from=5-4, to=5-7] \arrow["{\one_{\hat F}}", no head, from=3-7, to=5-7] \arrow["{\exists \eta_i}", dotted, from=3-7, to=3-10] \arrow["{\exists \eta_j}", dashed, from=5-7, to=5-10] \arrow["{\phi_i}", curve={height=-30pt}, dotted, from=3-4, to=3-10] \arrow["{\phi_j}", curve={height=30pt}, dotted, from=5-4, to=5-10] \arrow["{\one_X}"{description}, no head, from=3-10, to=5-10] \arrow["{F(\wait)}"', shorten <=30pt, shorten >=30pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Applying the $\diagonal$ functor, we can view this as a simpler universal property in $\cat{A}^{\cat{I}}$, since the above data is precisely the data of a natural transformation: \begin{tikzcd} {} && {\diagonal F} \\ \\ F && {\diagonal X} \arrow["\Phi"', from=3-1, to=3-3] \arrow["\Psi", from=3-1, to=1-3] \arrow["{\exists ! \eta}"', dashed, from=3-3, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwXSxbMiwwLCJcXGRpYWdvbmFsIEYiXSxbMCwyLCJGIl0sWzIsMiwiXFxkaWFnb25hbCBYIl0sWzIsMywiXFxQaGkiLDJdLFsyLDEsIlxcUHNpIl0sWzMsMSwiXFxleGlzdHMgISBcXGV0YSIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) That is, the functor $\diagonal \hat F$ is equipped with structure maps $\Phi: F\to \diagonal \hat{F}$ which assemble into a natural transformation (i.e. a morphism in $\cat{A}^{\cat{I}}$) such that any other natural transformation from $F$ to a diagonal object $\diagonal X$ produces a unique natural transformation $\eta: \diagonal X\to \diagonal F$. This provides exactly the data needed to specify $\tau$: \[ \tau_{FX}: \Hom_{\cat{A}}(\hat{F}, X) &\mapsvia{\sim} \Hom_{\cat{A}^{\cat{I}} }(F, \diagonal X) \\ \qty{ \hat{F} \mapsvia{f} X }&\mapsto \qty{ F \mapsvia{\Psi} \diagonal \hat{F} \mapsvia{\diagonal(f)} \diagonal X } ,\] i.e. we take the image $\Delta(f)$ and pre-compose with the structure morphism $\Psi$. ::: :::{.claim} This is a bijection of sets. ::: :::{.proof title="?"} This is surjective by the universal property: any morphism $F \mapsvia{g} \diagonal X$ in $\cat{A}^{\cat{I}}$ factors through $\diagonal \hat{F}$, and so all such morphisms are of this form. \todo[inline]{Todo: showing surjectivity or constructing inverse.} ::: :::{.claim} This is a natural isomorphism, i.e. for all \( X \mapsvia{f} Y \in\Mor(\cat{A}) \) and all \( F \mapsvia{\eta} G \in \Mor(\cat{A}^{\cat{I}}) \), there is a commuting diagram \begin{tikzcd} {\Hom(\hat{G}, X)} && {\Hom(\hat{F}, X)} && {\Hom(\hat{F}, Y)} \\ \\ {\Hom(G, \diagonal X)} && {\Hom(F, \diagonal X)} && {\Hom(F, \diagonal Y)} \arrow["{\tau_{GX}}", from=1-1, to=3-1] \arrow["{\tau_{FX}}", from=1-3, to=3-3] \arrow["{\tau_{FY}}", from=1-5, to=3-5] \arrow["{\hat\eta_*}", from=1-1, to=1-3] \arrow["{f_*}", from=1-3, to=1-5] \arrow["{\eta_*}", from=3-1, to=3-3] \arrow["{(\diagonal f)_*}", from=3-3, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJcXEhvbShcXGhhdHtGfSwgWCkiXSxbNCwwLCJcXEhvbShcXGhhdHtGfSwgWSkiXSxbMCwwLCJcXEhvbShcXGhhdHtHfSwgWCkiXSxbMCwyLCJcXEhvbShHLCBcXGRpYWdvbmFsIFgpIl0sWzIsMiwiXFxIb20oRiwgXFxkaWFnb25hbCBYKSJdLFs0LDIsIlxcSG9tKEYsIFxcZGlhZ29uYWwgWSkiXSxbMiwzLCJcXHRhdV97R1h9Il0sWzAsNCwiXFx0YXVfe0ZYfSJdLFsxLDUsIlxcdGF1X3tGWX0iXSxbMiwwLCJcXGhhdFxcZXRhXyoiXSxbMCwxLCJmXyoiXSxbMyw0LCJcXGV0YV8qIl0sWzQsNSwiKFxcZGlhZ29uYWwgZilfKiJdXQ==) ::: :::{.proof title="?"} \todo[inline]{Show that the diagrams above always commute.} ::: :::{.claim} If \( \cat{A} \) is abelian and $\cat{I}$ is an index category such that $\colim_{i\in \cat{I}}F(i)$ exists for all $F\in \cat{A}^{\cat{I}}$, then the functor $\colim: \cat{A}^{\cat{I}} \to \cat{A}$ is right-exact. ::: :::{.proof title="Sketch"} A sketch of the proof proceeds by showing every right adjoint is left-exact: - Since $\Hom(LA, \wait)$ 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 $\Hom(\wait, RB)$. - This sequence is exact, so applying Yoneda yields an exact sequence \[ 0\to RB' \to RB \to RB'' ,\] making $R$ left-exact. Finally, if $L$ is a left adjoint out of $\cat{A}$, then $L\op$ is a right adjoint out of $\cat{A}\op$. Thus $L\op$ is left-exact by the above argument, making $L$ right-exact. ::: :::{.claim} Let $\cat{I} \da \qty{ \bullet \leftarrow \bullet \rightarrow\bullet}$ and define the pushout as $\colim: \cat{A}^\cat{I} \to \cat{A}$. Then taking $\cat{A} \da \Ab$, the pushout does not define an exact functor $\cat{A}^\cat{I}\to \cat{A}$. ::: :::{.proof title="?"} We proceed by constructing a counterexample. Unwinding definitions, we first note that an exact sequence of objects in $\cat{A}^\cat{I}$ 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: \begin{tikzcd} 0 & 0 &&& 0 \\ 0 & \textcolor{rgb,255:red,92;green,92;blue,214}{P_1} & \textcolor{rgb,255:red,92;green,92;blue,214}{B_1} &&& \textcolor{rgb,255:red,92;green,214;blue,92}{X_1} \\ & \textcolor{rgb,255:red,92;green,92;blue,214}{A_1} & \textcolor{rgb,255:red,92;green,92;blue,214}{P_2} & \textcolor{rgb,255:red,92;green,92;blue,214}{B_2} &&& \textcolor{rgb,255:red,92;green,214;blue,92}{X_2} \\ && \textcolor{rgb,255:red,92;green,92;blue,214}{A_2} & \textcolor{rgb,255:red,92;green,92;blue,214}{P_3} & \textcolor{rgb,255:red,92;green,92;blue,214}{B_3} && {} & \textcolor{rgb,255:red,92;green,214;blue,92}{X_3} \\ &&& \textcolor{rgb,255:red,92;green,92;blue,214}{A_3} & 0 & 0 &&& 0 \\ &&&& 0 \arrow[from=1-2, to=2-3] \arrow[from=2-3, to=3-4] \arrow[from=1-1, to=2-2] \arrow[from=2-1, to=3-2] \arrow[from=2-2, to=3-3] \arrow[from=3-2, to=4-3] \arrow[from=4-3, to=5-4] \arrow[from=3-3, to=4-4] \arrow[from=3-4, to=4-5] \arrow[from=4-4, to=5-5] \arrow[from=5-4, to=6-5] \arrow[from=4-5, to=5-6] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-2, to=2-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=3-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=4-4, to=4-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=4-4, to=5-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=4-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-2, to=3-2] \arrow[from=1-5, to=2-6] \arrow[from=2-6, to=3-7] \arrow[from=3-7, to=4-8] \arrow[from=4-8, to=5-9] \arrow[shorten <=16pt, shorten >=24pt, Rightarrow, from=3-4, to=3-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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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 $f_i$: \begin{tikzcd}[column sep=1.5em] 0 & 0 &&& 0 \\ 0 & \textcolor{rgb,255:red,92;green,92;blue,214}{P_1} & \textcolor{rgb,255:red,92;green,92;blue,214}{B_1} &&& \textcolor{rgb,255:red,92;green,214;blue,92}{\coker f_1} \\ & \textcolor{rgb,255:red,92;green,92;blue,214}{0} & \textcolor{rgb,255:red,92;green,92;blue,214}{P_2} & \textcolor{rgb,255:red,92;green,92;blue,214}{B_2} &&& \textcolor{rgb,255:red,92;green,214;blue,92}{\coker f_2} \\ && \textcolor{rgb,255:red,92;green,92;blue,214}{0} & \textcolor{rgb,255:red,92;green,92;blue,214}{P_3} & \textcolor{rgb,255:red,92;green,92;blue,214}{B_3} && {} & \textcolor{rgb,255:red,92;green,214;blue,92}{\coker f_3} \\ &&& \textcolor{rgb,255:red,92;green,92;blue,214}{0} & 0 & 0 &&& 0 \\ &&&& 0 \arrow[from=1-2, to=2-3] \arrow[from=2-3, to=3-4] \arrow[from=1-1, to=2-2] \arrow[from=2-1, to=3-2] \arrow[from=2-2, to=3-3] \arrow[from=3-2, to=4-3] \arrow[from=4-3, to=5-4] \arrow[from=3-3, to=4-4] \arrow[from=3-4, to=4-5] \arrow[from=4-4, to=5-5] \arrow[from=5-4, to=6-5] \arrow[from=4-5, to=5-6] \arrow["{f_1}", color={rgb,255:red,92;green,92;blue,214}, from=2-2, to=2-3] \arrow["{f_2}", color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=3-4] \arrow["{f_3}", color={rgb,255:red,92;green,92;blue,214}, from=4-4, to=4-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=4-4, to=5-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=4-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-2, to=3-2] \arrow[from=1-5, to=2-6] \arrow[from=2-6, to=3-7] \arrow[from=3-7, to=4-8] \arrow[from=4-8, to=5-9] \arrow[shorten <=18pt, shorten >=27pt, Rightarrow, from=3-4, to=3-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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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: \begin{tikzcd} 0 & {\ker f_1} & {\ker f_2} & \textcolor{rgb,255:red,92;green,214;blue,92}{\ker f_3} \\ 0 & {P_1} & {P_1} & {P_1} & 0 \\ \\ 0 & {P_1} & {P_1} & {P_1} & 0 \\ & \textcolor{rgb,255:red,214;green,92;blue,92}{\coker f_1} & \textcolor{rgb,255:red,214;green,92;blue,92}{\coker f_2} & {\coker f_3} & 0 \\ & {} & {} & {} & {} \arrow["{f_1}", from=2-2, to=4-2] \arrow["{f_2}", from=2-3, to=4-3] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=4-1, to=4-2] \arrow[from=4-2, to=4-3] \arrow[from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=2-4, to=2-5] \arrow["{f_3}", from=2-4, to=4-4] \arrow[from=1-3, to=1-4] \arrow[from=1-2, to=1-3] \arrow[from=1-1, to=1-2] \arrow[draw={rgb,255:red,92;green,214;blue,92}, from=1-4, to=5-2, in=180, out=0] \arrow[draw={rgb,255:red,214;green,92;blue,92}, from=5-2, to=5-3] \arrow[from=5-3, to=5-4] \arrow[from=5-4, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Here we know that the map involved in the red terms $\coker f_1 \to \coker f_2$ 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. :::