Lemma 13.16.4. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset containing $0$ such that every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$. Let $a \in \mathbf{Z}$.

1. Given $K^\bullet$ with $K^ n = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet$ with $K^ n \to I^ n$ injective and $I^ n \in \mathcal{I}$ for all $n$ and $I^ n = 0$ for $n < a$,

2. Given $K^\bullet$ with $H^ n(K^\bullet ) = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet$ with $I^ n \in \mathcal{I}$ and $I^ n = 0$ for $n < a$.

Proof. Proof of part (1). Consider the following induction hypothesis $IH_ n$: There are $I^ j \in \mathcal{I}$, $j \leq n$ almost all zero, maps $d^ j : I^ j \to I^{j + 1}$ for $j < n$ and injective maps $\alpha ^ j : K^ j \to I^ j$ for $j \leq n$ such that the diagram

$\xymatrix{ \ldots \ar[r] & K^{n - 1} \ar[d]^\alpha \ar[r] & K^ n \ar[d]^\alpha \ar[r] & K^{n + 1} \ar[r] & \ldots \\ \ldots \ar[r] & I^{n - 1} \ar[r] & I^ n & & }$

is commutative, such that $d^ j \circ d^{j - 1} = 0$ for $j < n$ and such that $\alpha$ induces isomorphisms $H^ j(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ j)/\mathop{\mathrm{Im}}(d^{j - 1})$ for $j < n$. Note that this implies

13.16.4.1
\begin{equation} \label{derived-equation-new-star} \alpha (\mathop{\mathrm{Im}}(d_ K^{n - 1})) \subset \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)) \cap \mathop{\mathrm{Im}}(d^{n - 1}) \subset \alpha (K^ n) \cap \mathop{\mathrm{Im}}(d^{n - 1}). \end{equation}

If these inclusions are not equalities, then choose an injection

$I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1}) \longrightarrow I$

with $I \in \mathcal{I}$. Denote $\alpha ' : K^ n \to I$ the map obtained by composing $\alpha \oplus 1 : K^ n \to I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ with the displayed injection. Denote $d' : I^{n - 1} \to I$ the composition $I^{n - 1} \to I^ n \to I$ of $d^{n - 1}$ by the inclusion of the first summand. Then $\alpha '(K^ n) \cap \mathop{\mathrm{Im}}(d') = \alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$ simply because the intersection of $\alpha '(K^ n)$ with the first summand of $I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ is equal to $\alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$. Hence, after replacing $I^ n$ by $I$, $\alpha$ by $\alpha '$ and $d^{n - 1}$ by $d'$ we may assume that we have equality in Equation (13.16.4.1). Once this is the case consider the solid diagram

$\xymatrix{ K^ n/\mathop{\mathrm{Ker}}(d_ K^ n) \ar[r] \ar[d] & K^{n + 1} \ar@{..>}[d] \\ I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n))) \ar@{..>}[r] & M }$

The horizontal arrow is injective by fiat and the vertical arrow is injective as we have equality in (13.16.4.1). Hence the push-out $M$ of this diagram contains both $K^{n + 1}$ and $I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)))$ as subobjects. Choose an injection $M \to I^{n + 1}$ with $I^{n + 1} \in \mathcal{I}$. By construction we get $d^ n : I^ n \to I^{n + 1}$ and an injective map $\alpha ^{n + 1} : K^{n + 1} \to I^{n + 1}$. The equality in Equation (13.16.4.1) and the construction of $d^ n$ guarantee that $\alpha : H^ n(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ n)/\mathop{\mathrm{Im}}(d^{n - 1})$ is an isomorphism. In other words $IH_{n + 1}$ holds.

We finish the proof of by the following observations. First we note that $IH_ n$ is true for $n = a$ since we can just take $I^ j = 0$ for $j < a$ and $K^ a \to I^ a$ an injection of $K^ a$ into an element of $\mathcal{I}$. Next, we note that in the proof of $IH_ n \Rightarrow IH_{n + 1}$ we only modified the object $I^ n$, the map $d^{n - 1}$ and the map $\alpha ^ n$. Hence we see that proceeding by induction we produce a complex $I^\bullet$ with $I^ n = 0$ for $n < a$ consisting of objects from $\mathcal{I}$, and a termwise injective quasi-isomorphism $\alpha : K^\bullet \to I^\bullet$ as desired.

Proof of part (2). The assumption implies that the morphism $K^\bullet \to \tau _{\geq a}K^\bullet$ (Homology, Section 12.14) is a quasi-isomorphism. Apply part (1) to find $\tau _{\geq a}K^\bullet \to I^\bullet$. The composition $K^\bullet \to I^\bullet$ is the desired quasi-isomorphism. $\square$

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