
14.27 Homotopies in abelian categories

Let $\mathcal{A}$ be an abelian category. Let $U$, $V$ be simplicial objects of $\mathcal{A}$. Let $a, b : U \to V$ be morphisms. Further, suppose that $h : U \times \Delta [1] \to V$ is a homotopy connecting $a$ and $b$. Consider the two morphisms of chain complexes $s(a), s(b) : s(U) \longrightarrow s(V)$. Using the notation introduced above Lemma 14.26.2 we define

$s(h)_ n : U_ n \longrightarrow V_{n + 1}$

by the formula

14.27.0.1
$$\label{simplicial-equation-homotopy-to-homotopy} s(h)_ n = \sum \nolimits _{i = 0}^ n (-1)^{i + 1} h_{n + 1, i + 1} \circ s^ n_ i.$$

Let us compute $d_{n + 1} \circ s(h)_ n + s(h)_{n - 1} \circ d_ n$. We first compute

\begin{eqnarray*} d_{n + 1} \circ s(h)_ n & = & \sum \nolimits _{j = 0}^{n + 1} \sum \nolimits _{i = 0}^ n (-1)^{j + i + 1} d^{n + 1}_ j \circ h_{n + 1, i + 1} \circ s^ n_ i \\ & = & \sum \nolimits _{1 \leq i + 1 \leq j \leq n + 1} (-1)^{j + i + 1} h_{n, i + 1} \circ d^{n + 1}_ j \circ s^ n_ i \\ & & + \sum \nolimits _{n \geq i \geq j \geq 0} (-1)^{i + j + 1} h_{n, i} \circ d^{n + 1}_ j \circ s^ n_ i \\ & = & \sum \nolimits _{1 \leq i + 1 < j \leq n + 1} (-1)^{j + i + 1} h_{n, i + 1} \circ s^{n - 1}_ i \circ d^ n_{j - 1} \\ & & + \sum \nolimits _{1 \leq i + 1 = j \leq n + 1} (-1)^{j + i + 1} h_{n, i + 1} \\ & & + \sum \nolimits _{n \geq i = j \geq 0} (-1)^{i + j + 1} h_{n, i} \\ & & + \sum \nolimits _{n \geq i > j \geq 0} (-1)^{i + j + 1} h_{n, i} \circ s^{n - 1}_{i - 1} \circ d^ n_ j \end{eqnarray*}

We leave it to the reader to see that the first and the last of the four sums cancel exactly against all the terms of

$s(h)_{n - 1} \circ d_ n = \sum _{i = 0}^{n - 1} \sum _{j = 0}^ n (-1)^{i + 1 + j} h_{n, i + 1} \circ s^{n - 1}_ i \circ d^ n_ j.$

Hence we obtain

\begin{eqnarray*} d_{n + 1} \circ s(h)_ n + s(h)_{n - 1} \circ d_ n & = & \sum _{j = 1}^{n + 1} (-1)^{2j} h_{n, j} + \sum _{i = 0}^ n (-1)^{2i + 1} h_{n, i} \\ & = & h_{n, n + 1} - h_{n , 0} \\ & = & a_ n - b_ n \end{eqnarray*}

Thus we've proved part of the following lemma.

Lemma 14.27.1. Let $\mathcal{A}$ be an abelian category. Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{A}$. If $a$, $b$ are homotopic, then $s(a), s(b) : s(U) \to s(V)$, and $N(a), N(b) : N(U) \to N(V)$ are homotopic maps of chain complexes.

Proof. The part about $s(a)$ and $s(b)$ is clear from the calculation above the lemma. On the other hand, if follows from Lemma 14.23.6 that $N(a)$, $N(b)$ are compositions

$N(U) \to s(U) \to s(V) \to N(V)$

where we use $s(a)$, $s(b)$ in the middle. Hence the assertion follows from Homology, Lemma 12.12.1. $\square$

Lemma 14.27.2. Let $\mathcal{A}$ be an abelian category. Let $a : U \to V$ be a morphism of simplicial objects of $\mathcal{A}$. If $a$ is a homotopy equivalence, then $s(a) : s(U) \to s(V)$, and $N(a) : N(U) \to N(V)$ are homotopy equivalences of chain complexes.

Proof. Omitted. See Lemma 14.27.1 above. $\square$

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