**Proof.**
Proof of part (1). Consider the following induction hypothesis $IH_ n$: There are $P^ j \in \mathcal{P}$, $j \geq n$, with $P^ j = 0$ for $j > a$, maps $d^ j : P^ j \to P^{j + 1}$ for $j \geq n$, and surjective maps $\alpha ^ j : P^ j \to K^ j$ for $j \geq n$ such that the diagram

\[ \xymatrix{ & & P^ n \ar[d]^\alpha \ar[r] & P^{n + 1} \ar[d]^\alpha \ar[r] & P^{n + 2} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & K^{n + 2} \ar[r] & \ldots } \]

is commutative, such that $d^{j + 1} \circ d^ j = 0$ for $j \geq n$, such that $\alpha $ induces isomorphisms $\mathop{\mathrm{Ker}}(d^ j)/\mathop{\mathrm{Im}}(d^{j - 1}) \to H^ j(K^\bullet )$ for $j > n$, and such that $\alpha : \mathop{\mathrm{Ker}}(d^ n) \to \mathop{\mathrm{Ker}}(d_ K^ n)$ is surjective. Then we choose a surjection

\[ P^{n - 1} \longrightarrow K^{n - 1} \times _{K^ n} \mathop{\mathrm{Ker}}(d^ n) = K^{n - 1} \times _{\mathop{\mathrm{Ker}}(d_ K^ n)} \mathop{\mathrm{Ker}}(d^ n) \]

with $P^{n - 1}$ in $\mathcal{P}$. This allows us to extend the diagram above to

\[ \xymatrix{ & P^{n - 1} \ar[d]^\alpha \ar[r] & P^ n \ar[d]^\alpha \ar[r] & P^{n + 1} \ar[d]^\alpha \ar[r] & P^{n + 2} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & K^{n + 2} \ar[r] & \ldots } \]

The reader easily checks that $IH_{n - 1}$ holds with this choice.

We finish the proof of (1) as follows. First we note that $IH_ n$ is true for $n = a + 1$ since we can just take $P^ j = 0$ for $j > a$. Hence we see that proceeding by descending induction we produce a complex $P^\bullet $ with $P^ n = 0$ for $n > a$ consisting of objects from $\mathcal{P}$, and a termwise surjective quasi-isomorphism $\alpha : P^\bullet \to K^\bullet $ as desired.

Proof of part (2). The assumption implies that the morphism $\tau _{\leq a}K^\bullet \to K^\bullet $ (Homology, Section 12.15) is a quasi-isomorphism. Apply part (1) to find $P^\bullet \to \tau _{\leq a}K^\bullet $. The composition $P^\bullet \to K^\bullet $ is the desired quasi-isomorphism.
$\square$

## Comments (2)

Comment #8396 by ElĂas Guisado on

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