Lemma 46.8.6. Let $A$ be a ring.

1. Any $A$-module has a pure projective resolution.

Let $M \to N$ be a map of $A$-modules. Let $P_\bullet \to M$ be a pure projective resolution and let $N_\bullet \to N$ be a universally exact resolution.

1. There exists a map of complexes $P_\bullet \to N_\bullet$ inducing the given map

$M = \mathop{\mathrm{Coker}}(P_1 \to P_0) \to \mathop{\mathrm{Coker}}(N_1 \to N_0) = N$
2. two maps $\alpha , \beta : P_\bullet \to N_\bullet$ inducing the same map $M \to N$ are homotopic.

Proof. Part (1) follows immediately from Lemma 46.8.2. Before we prove (2) and (3) note that by Lemma 46.8.4 we can split the universally exact complex $N_\bullet \to N \to 0$ into universally exact short exact sequences $0 \to K_0 \to N_0 \to N \to 0$ and $0 \to K_ i \to N_ i \to K_{i - 1} \to 0$.

Proof of (2). Because $P_0$ is pure projective we can find a map $P_0 \to N_0$ lifting the map $P_0 \to M \to N$. We obtain an induced map $P_1 \to F_0 \to N_0$ which ends up in $K_0$. Since $P_1$ is pure projective we may lift this to a map $P_1 \to N_1$. This in turn induces a map $P_2 \to P_1 \to N_1$ which maps to zero into $N_0$, i.e., into $K_1$. Hence we may lift to get a map $P_2 \to N_2$. Repeat.

Proof of (3). To show that $\alpha , \beta$ are homotopic it suffices to show the difference $\gamma = \alpha - \beta$ is homotopic to zero. Note that the image of $\gamma _0 : P_0 \to N_0$ is contained in $K_0$. Hence we may lift $\gamma _0$ to a map $h_0 : P_0 \to N_1$. Consider the map $\gamma _1' = \gamma _1 - h_0 \circ d_{P, 1} : P_1 \to N_1$. By our choice of $h_0$ we see that the image of $\gamma _1'$ is contained in $K_1$. Since $P_1$ is pure projective may lift $\gamma _1'$ to a map $h_1 : P_1 \to N_2$. At this point we have $\gamma _1 = h_0 \circ d_{F, 1} + d_{N, 2} \circ h_1$. Repeat. $\square$

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