The Stacks project

Lemma 13.18.6. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

\[ \xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^\beta \\ I^\bullet } \]

where $I^\bullet $ is bounded below and consists of injective objects, and $\alpha $ is a quasi-isomorphism.

  1. There exists a map of complexes $\beta $ making the diagram commute up to homotopy.

  2. If $\alpha $ is injective in every degree then we can find a $\beta $ which makes the diagram commute.

Proof. The “correct” proof of part (1) is explained in Remark 13.18.5. We also give a direct proof here.

We first show that (2) implies (1). Namely, let $\tilde\alpha : K \to \tilde L^\bullet $, $\pi $, $s$ be as in Lemma 13.9.6. Since $\tilde\alpha $ is injective by (2) there exists a morphism $\tilde\beta : \tilde L^\bullet \to I^\bullet $ such that $\gamma = \tilde\beta \circ \tilde\alpha $. Set $\beta = \tilde\beta \circ s$. Then we have

\[ \beta \circ \alpha = \tilde\beta \circ s \circ \pi \circ \tilde\alpha \sim \tilde\beta \circ \tilde\alpha = \gamma \]

as desired.

Assume that $\alpha : K^\bullet \to L^\bullet $ is injective. Suppose we have already defined $\beta $ in all degrees $\leq n - 1$ compatible with differentials and such that $\gamma ^ j = \beta ^ j \circ \alpha ^ j$ for all $j \leq n - 1$. Consider the commutative solid diagram

\[ \xymatrix{ K^{n - 1} \ar[r] \ar@/_2pc/[dd]_\gamma \ar[d]^\alpha & K^ n \ar@/^2pc/[dd]^\gamma \ar[d]^\alpha \\ L^{n - 1} \ar[r] \ar[d]^\beta & L^ n \ar@{-->}[d] \\ I^{n - 1} \ar[r] & I^ n } \]

Thus we see that the dotted arrow is prescribed on the subobjects $\alpha (K^ n)$ and $d^{n - 1}(L^{n - 1})$. Moreover, these two arrows agree on $\alpha (d^{n - 1}(K^{n - 1}))$. Hence if
\begin{equation} \label{derived-equation-qis} \alpha (d^{n - 1}(K^{n - 1})) = \alpha (K^ n) \cap d^{n - 1}(L^{n - 1}) \end{equation}

then these morphisms glue to a morphism $\alpha (K^ n) + d^{n - 1}(L^{n - 1}) \to I^ n$ and, using the injectivity of $I^ n$, we can extend this to a morphism from all of $L^ n$ into $I^ n$. After this by induction we get the morphism $\beta $ for all $n$ simultaneously (note that we can set $\beta ^ n = 0$ for all $n \ll 0$ since $I^\bullet $ is bounded below – in this way starting the induction).

It remains to prove the equality ( The reader is encouraged to argue this for themselves with a suitable diagram chase. Nonetheless here is our argument. Note that the inclusion $\alpha (d^{n - 1}(K^{n - 1})) \subset \alpha (K^ n) \cap d^{n - 1}(L^{n - 1})$ is obvious. Take an object $T$ of $\mathcal{A}$ and a morphism $x : T \to L^ n$ whose image is contained in the subobject $\alpha (K^ n) \cap d^{n - 1}(L^{n - 1})$. Since $\alpha $ is injective we see that $x = \alpha \circ x'$ for some $x' : T \to K^ n$. Moreover, since $x$ lies in $d^{n - 1}(L^{n - 1})$ we see that $d^ n \circ x = 0$. Hence using injectivity of $\alpha $ again we see that $d^ n \circ x' = 0$. Thus $x'$ gives a morphism $[x'] : T \to H^ n(K^\bullet )$. On the other hand the corresponding map $[x] : T \to H^ n(L^\bullet )$ induced by $x$ is zero by assumption. Since $\alpha $ is a quasi-isomorphism we conclude that $[x'] = 0$. This of course means exactly that the image of $x'$ is contained in $d^{n - 1}(K^{n - 1})$ and we win. $\square$

Comments (2)

Comment #8416 by on

It took me a while to find out how the map is defined. Here is a proof by chasing elements: it suffices to show that the composite vanishes. Let . Since is an isomorphism, there are and such that . Hence, the image of along equals and vanishes by commutativity of the solid diagram, for .

There are also:

  • 3 comment(s) on Section 13.18: Injective resolutions

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