Lemma 13.18.7. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

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

where $I^\bullet$ is bounded below and consists of injective objects, and $\alpha$ is a quasi-isomorphism. Any two morphisms $\beta _1, \beta _2$ making the diagram commute up to homotopy are homotopic.

Proof. This follows from Remark 13.18.5. We also give a direct argument here.

Let $\tilde\alpha : K \to \tilde L^\bullet$, $\pi$, $s$ be as in Lemma 13.9.6. If we can show that $\beta _1 \circ \pi$ is homotopic to $\beta _2 \circ \pi$, then we deduce that $\beta _1 \sim \beta _2$ because $\pi \circ s$ is the identity. Hence we may assume $\alpha ^ n : K^ n \to L^ n$ is the inclusion of a direct summand for all $n$. Thus we get a short exact sequence of complexes

$0 \to K^\bullet \to L^\bullet \to M^\bullet \to 0$

which is termwise split and such that $M^\bullet$ is acyclic. We choose splittings $L^ n = K^ n \oplus M^ n$, so we have $\beta _ i^ n : K^ n \oplus M^ n \to I^ n$ and $\gamma ^ n : K^ n \to I^ n$. In this case the condition on $\beta _ i$ is that there are morphisms $h_ i^ n : K^ n \to I^{n - 1}$ such that

$\gamma ^ n - \beta _ i^ n|_{K^ n} = d \circ h_ i^ n + h_ i^{n + 1} \circ d$

Thus we see that

$\beta _1^ n|_{K^ n} - \beta _2^ n|_{K^ n} = d \circ (h_1^ n - h_2^ n) + (h_1^{n + 1} - h_2^{n + 1}) \circ d$

Consider the map $h^ n : K^ n \oplus M^ n \to I^{n - 1}$ which equals $h_1^ n - h_2^ n$ on the first summand and zero on the second. Then we see that

$\beta _1^ n - \beta _2^ n - (d \circ h^ n + h^{n + 1}) \circ d$

is a morphism of complexes $L^\bullet \to I^\bullet$ which is identically zero on the subcomplex $K^\bullet$. Hence it factors as $L^\bullet \to M^\bullet \to I^\bullet$. Thus the result of the lemma follows from Lemma 13.18.4. $\square$

Comment #5525 by Wen-Wei LI on

Extra ) in the last displayed formula of this proof.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).