Lemma 13.18.4 (013R)—The Stacks project

Lemma 13.18.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be an acyclic complex. Let $I^\bullet$ be bounded below and consisting of injective objects. Any morphism $K^\bullet \to I^\bullet$ is homotopic to zero.

Proof. Let $\alpha : K^\bullet \to I^\bullet$ be a morphism of complexes. Note that $\alpha ^ j = 0$ for $j \ll 0$ as $I^\bullet$ is bounded below. In particular, we can find an $n$ such that there exist $h^ j : K^ j \to I^{j - 1}$ for $j \leq n$ such that $\alpha ^ j = d^{j - 1} \circ h^ j + h^{j + 1} \circ d^ j$ for $j < n$ . We will show that there exists a morphism $h^{n + 1} : K^{n + 1} \to I^ n$ such that $\alpha ^ n = d^{n - 1} \circ h^ n + h^{n + 1} \circ d^ n$ . Note that

$\begin{align*} (\alpha ^ n - d^{n - 1} \circ h^{n - 1}) \circ d^{n - 1} & = \alpha ^{n - 1} \circ d^{n - 1} - d^{n - 1} \circ h^{n - 1} \circ d^{n - 1} \\ & = d^{n - 1} \circ \alpha ^{n - 1} - d^{n - 1} \circ h^{n - 1} \circ d^{n - 1} \\ & = d^{n - 1} \circ (d^{n - 2} \circ h^{n - 1} + h^ n \circ d^{n - 1}) - d^{n - 1} \circ h^{n - 1} \circ d^{n - 1} \\ & = 0 \end{align*}$

Since $K^\bullet$ is acyclic we have $d^{n - 1}(K^{n - 1}) = \mathop{\mathrm{Ker}}(K^ n \to K^{n + 1})$ . Hence we can think of $\alpha ^ n - d^{n - 1} \circ h^{n - 1}$ as a map into $I^ n$ defined on the subobject $\mathop{\mathrm{Im}}(K^ n \to K^{n + 1})$ of $K^{n + 1}$ . By injectivity of the object $I^ n$ we can extend this to a map $h^{n + 1} : K^{n + 1} \to I^ n$ . With this choice the reader checks that we indeed have $\alpha ^ n = d^{n - 1} \circ h^ n + h^{n + 1} \circ d^ n$ .

By induction on $n$ we conclude we can find $h = (h^ j)_{j \in \mathbf{Z}}$ which forms a homotopy between $\alpha$ and $0$ as desired. $\square$

Comments (3)

Comment #8414 by Elías Guisado on May 19, 2023 at 03:54

I think what the proof does might not be enough to conclude that $K^\bullet\to I^\bullet$ is homotopic to zero, https://mathoverflow.net/questions/185366/is-such-a-map-null-homotopic

Comment #8415 by Elías Guisado on May 21, 2023 at 14:36

The correct proof is not much different from the actual one: suppose there is $n$ and that, for each $i\leq n$ , there are maps $h^i:K^i\to I^{i-1}$ such that $\alpha^i=d^{i-1}h^i+h^{i+1}d^i$ for all $i<n$ . We are going to construct a map $h^{n+1}:K^{n+1}\to I^n$ such that $\alpha^n-d^{n-1}h^n=h^{n+1}d^n$ . For this, note that $\begin{gather} (\alpha^n-d^{n-1}h^n)d^{n-1}=\alpha^nd^{n-1}-d^{n-1}h^nd^{n-1}=\\ \alpha^nd^{n-1}-d^{n-1}(\alpha^{n-1}-d^{n-2}h^{n-1})=\alpha^nd^{n-1}-d^{n-1}\alpha^{n-1}=0. \end{gather}$ Hence, $0=(\alpha^n-d^{n-1}h^n)|_{d^{n-1}(K^{n-1})}$ , and we can reuse the already existing argument.

Comment #9039 by Stacks project on May 03, 2024 at 12:28

@#8414: I think this mathoverflow post discusses a different thing because here the proof as it is written now shows there exists a homotopy to go from a map zero in degrees $< n$ to a map zero in degrees $< n + 1$ where moreover the homotopy is zero except for the map $h : K^{n + 1} \to I^n$ and that clearly does work (in general). Anyway, I have changed it as you suggested here so the point I just made is now moot (or will be when the site is updated). Thanks.

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