The Stacks project

Proof. Omitted. For the second statement use $L^\bullet = \tau _{\geq n}K^\bullet $ for some $n \ll 0$. See Homology, Section 12.15 for the definition of the truncation $\tau _{\geq n}$. $\square$

Proof. Proof of (1). First choose an injection $A \to I^0$ of $A$ into an injective object of $\mathcal{A}$. Next, choose an injection $I_0/A \to I^1$ into an injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose an injection $I^1/\mathop{\mathrm{Im}}(d^0) \to I^2$ into an injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. By Lemma 13.18.2 part (2) follows from part (3). Part (3) is a special case of Lemma 13.15.5. $\square$

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

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. $\square$

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

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

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

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 (13.18.6.1). 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. $\square$

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

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

Thus we see that

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

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. $\square$

Proof. Let $a$ be an element of the right hand side. We may represent $a = \gamma \alpha ^{-1}$ where $\alpha : K^\bullet \to L^\bullet $ is a quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet $ is a map of complexes. By Lemma 13.18.6 we can find a morphism $\beta : L^\bullet \to I^\bullet $ such that $\beta \circ \alpha $ is homotopic to $\gamma $. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet $ such that there exists a quasi-isomorphism $\alpha : K^\bullet \to L^\bullet $ with $\beta \circ \alpha $ homotopic to zero. Then Lemma 13.18.7 shows that $\beta $ is homotopic to zero also, i.e., $b = 0$. $\square$

Proof. Step 1. Choose an injective resolution $A^\bullet \to I^\bullet $ (see Lemma 13.18.3) or use the given one. Recall that $\text{Comp}^{+}(\mathcal{A})$ is an abelian category, see Homology, Lemma 12.13.9. Hence we may form the pushout along the map $A^\bullet \to I^\bullet $ to get

Because of the $5$-lemma and the last assertion of Homology, Lemma 12.13.12 the map $B^\bullet \to E^\bullet $ is a quasi-isomorphism. Note that the lower short exact sequence is termwise split, see Homology, Lemma 12.27.2. Hence it suffices to prove the lemma when $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ is termwise split.

Step 2. Choose splittings. In other words, write $B^ n = A^ n \oplus C^ n$. Denote $\delta : C^\bullet \to A^\bullet [1]$ the morphism as in Homology, Lemma 12.14.10. Choose injective resolutions $f_1 : A^\bullet \to I_1^\bullet $ and $f_3 : C^\bullet \to I_3^\bullet $. (If $A^\bullet $ is a complex of injectives, then use $I_1^\bullet = A^\bullet $.) We may assume $f_3$ is injective in every degree. By Lemma 13.18.6 we may find a morphism $\delta ' : I_3^\bullet \to I_1^\bullet [1]$ such that $\delta ' \circ f_3 = f_1[1] \circ \delta $ (equality of morphisms of complexes). Set $I_2^ n = I_1^ n \oplus I_3^ n$. Define

and define the maps $B^ n \to I_2^ n$ to be given as the sum of the maps $A^ n \to I_1^ n$ and $C^ n \to I_3^ n$. Everything is clear. $\square$