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. Assume that $\alpha ^ j = 0$ for $j < n$. We will show that there exists a morphism $h : K^{n + 1} \to I^ n$ such that $\alpha ^ n = h \circ d$. Thus $\alpha$ will be homotopic to the morphism of complexes $\beta$ defined by

$\beta ^ j = \left\{ \begin{matrix} 0 & \text{if} & j \leq n \\ \alpha ^{n + 1} - d \circ h & \text{if} & j = n + 1 \\ \alpha ^ j & \text{if} & j > n + 1 \end{matrix} \right.$

This will clearly prove the lemma (by induction). To prove the existence of $h$ note that $\alpha ^ n|_{d^{n - 1}(K^{n - 1})} = 0$ since $\alpha ^{n - 1} = 0$. 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 $\alpha ^ n$ 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 $h : K^{n + 1} \to I^ n$ as desired. $\square$

Comment #8414 by on

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 on

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. 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 Hence, $0=(\alpha^n-d^{n-1}h^n)|_{d^{n-1}(K^{n-1})}$, and we can reuse the already existing argument.

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