Lemma 10.71.4. Let $R$ be a ring. Let $M \to N$ be a map of $R$-modules. Let $N_\bullet \to N$ be an arbitrary resolution. Let

\[ \ldots \to F_2 \to F_1 \to F_0 \to M \]

be a complex of $R$-modules where each $F_ i$ is a free $R$-module. Then

there exists a map of complexes $F_\bullet \to N_\bullet $ such that

\[ \xymatrix{ F_0 \ar[r] \ar[d] & M \ar[d] \\ N_0 \ar[r] & N } \]

is commutative, and

any two maps $\alpha , \beta : F_\bullet \to N_\bullet $ as in (1) are homotopic.

**Proof.**
Proof of (1). Because $F_0$ is free we can find a map $F_0 \to N_0$ lifting the map $F_0 \to M \to N$. We obtain an induced map $F_1 \to F_0 \to N_0$ which ends up in the image of $N_1 \to N_0$. Since $F_1$ is free we may lift this to a map $F_1 \to N_1$. This in turn induces a map $F_2 \to F_1 \to N_1$ which maps to zero into $N_0$. Since $N_\bullet $ is exact we see that the image of this map is contained in the image of $N_2 \to N_1$. Hence we may lift to get a map $F_2 \to N_2$. Repeat.

Proof of (2). 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 $\gamma _0 : F_0 \to N_0$ is contained in the image of $N_1 \to N_0$. Hence we may lift $\gamma _0$ to a map $h_0 : F_0 \to N_1$. Consider the map $\gamma _1' = \gamma _1 - h_0 \circ d_{F, 1}$. By our choice of $h_0$ we see that the image of $\gamma _1'$ is contained in the kernel of $N_1 \to N_0$. Since $N_\bullet $ is exact we may lift $\gamma _1'$ to a map $h_1 : F_1 \to N_2$. 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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