Lemma 10.71.5. Let $R$ be a ring. Let $M_1, M_2, N$ be $R$-modules. Suppose that $F_{\bullet }$ is a free resolution of the module $M_1$, and $G_{\bullet }$ is a free resolution of the module $M_2$. Let $\varphi : M_1 \to M_2$ be a module map. Let $\alpha : F_{\bullet } \to G_{\bullet }$ be a map of complexes inducing $\varphi $ on $M_1 = \mathop{\mathrm{Coker}}(d_{F, 1}) \to M_2 = \mathop{\mathrm{Coker}}(d_{G, 1})$, see Lemma 10.71.4. Then the induced maps

\[ H^ i(\alpha ) : H^ i(\mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N)) \longrightarrow H^ i(\mathop{\mathrm{Hom}}\nolimits _ R(G_{\bullet }, N)) \]

are independent of the choice of $\alpha $. If $\varphi $ is an isomorphism, so are all the maps $H^ i(\alpha )$. If $M_1 = M_2$, $F_\bullet = G_\bullet $, and $\varphi $ is the identity, so are all the maps $H_ i(\alpha )$.

**Proof.**
Another map $\beta : F_{\bullet } \to G_{\bullet }$ inducing $\varphi $ is homotopic to $\alpha $ by Lemma 10.71.4. Hence the maps $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) \to \mathop{\mathrm{Hom}}\nolimits _ R(G_\bullet , N)$ are homotopic. Hence the independence result follows from Lemma 10.71.3.

Suppose that $\varphi $ is an isomorphism. Let $\psi : M_2 \to M_1$ be an inverse. Choose $\beta : G_{\bullet } \to F_{\bullet }$ be a map inducing $\psi : M_2 = \mathop{\mathrm{Coker}}(d_{G, 1}) \to M_1 = \mathop{\mathrm{Coker}}(d_{F, 1})$, see Lemma 10.71.4. OK, and now consider the map $H^ i(\alpha ) \circ H^ i(\beta ) = H^ i(\alpha \circ \beta )$. By the above the map $H^ i(\alpha \circ \beta )$ is the *same* as the map $H^ i(\text{id}_{G_{\bullet }}) = \text{id}$. Similarly for the composition $H^ i(\beta ) \circ H^ i(\alpha )$. Hence $H^ i(\alpha )$ and $H^ i(\beta )$ are inverses of each other.
$\square$

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