Lemma 14.29.1 (01A2)—The Stacks project

Lemma 14.29.1. Let $\mathcal{A}$ be an abelian category. Let $A$ be a chain complex. Consider the covariant functor

$B \longmapsto \{ (a, b, h) \mid a, b : A \to B\text{ and }h\text{ a homotopy between }a, b \}$

There exists a chain complex $\diamond A$ such that $\mathop{\mathrm{Mor}}\nolimits _{\text{Ch}(\mathcal{A})}(\diamond A, -)$ is isomorphic to the displayed functor. The construction $A \mapsto \diamond A$ is functorial.

Proof. We set $\diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$ , and we define $d_{\diamond A, n}$ by the matrix

$d_{\diamond A, n} = \left( \begin{matrix} d_{A, n} & 0 & \text{id}_{A_{n - 1}} \\ 0 & d_{A, n} & -\text{id}_{A_{n - 1}} \\ 0 & 0 & -d_{A, n - 1} \end{matrix} \right) : A_ n \oplus A_ n \oplus A_{n - 1} \to A_{n - 1} \oplus A_{n - 1} \oplus A_{n - 2}$

If $\mathcal{A}$ is the category of abelian groups, and $(x, y, z) \in A_ n \oplus A_ n \oplus A_{n - 1}$ then $d_{\diamond A, n}(x, y, z) = (d_ n(x) + z, d_ n(y) - z, -d_{n - 1}(z))$ . It is easy to verify that $d^2 = 0$ . Clearly, there are two maps $\diamond a, \diamond b : A \to \diamond A$ (first summand and second summand), and a map $\diamond A \to A[-1]$ which give a short exact sequence

$0 \to A \oplus A \to \diamond A \to A[-1] \to 0$

which is termwise split. Moreover, there is a sequence of maps $\diamond h_ n : A_ n \to \diamond A_{n + 1}$ , namely the identity from $A_ n$ to the summand $A_ n$ of $\diamond A_{n + 1}$ , such that $\diamond h$ is a homotopy between $\diamond a$ and $\diamond b$ .

We conclude that any morphism $f : \diamond A \to B$ gives rise to a triple $(a, b, h)$ by setting $a = f \circ \diamond a$ , $b = f \circ \diamond b$ and $h_ n = f_{n + 1} \circ \diamond h_ n$ . Conversely, given a triple $(a, b, h)$ we get a morphism $f : \diamond A \to B$ by taking

$f_ n = (a_ n, b_ n, h_{n - 1}).$

To see that this is a morphism of chain complexes you have to do a calculation. We only do this in case $\mathcal{A}$ is the category of abelian groups: Say $(x, y, z) \in \diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$ . Then

$\begin{eqnarray*} f_{n - 1}(d_ n(x, y, z)) & = & f_{n - 1}(d_ n(x) + z, d_ n(y) - z, -d_{n - 1}(z)) \\ & = & a_ n(d_ n(x)) + a_ n(z) + b_ n(d_ n(y)) - b_ n(z) - h_{n - 2}(d_{n - 1}(z)) \end{eqnarray*}$

and

$\begin{eqnarray*} d_ n(f_ n(x, y, z) & = & d_ n(a_ n(x) + b_ n(y) + h_{n - 1}(z)) \\ & = & d_ n(a_ n(x)) + d_ n(b_ n(y)) + d_ n(h_{n - 1}(z)) \end{eqnarray*}$

which are the same by definition of a homotopy. $\square$

The Stacks project

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