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
Comments (1)
Comment #9942 by Tony on
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