Lemma 14.27.1. Let $\mathcal{A}$ be an additive category. Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{A}$. If $a$, $b$ are homotopic, then $s(a), s(b) : s(U) \to s(V)$ are homotopic maps of chain complexes. If $\mathcal{A}$ is abelian, then also $N(a), N(b) : N(U) \to N(V)$ are homotopic maps of chain complexes.
14.27 Homotopies in abelian categories
Let $\mathcal{A}$ be an additive 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 from $a$ to $b$. Let us prove the two morphisms of chain complexes $s(a), s(b) : s(U) \longrightarrow s(V)$ are homotopic in the sense of Homology, Section 12.13. Using the notation introduced in Section 14.26 we define
\[ s(h)_ n : U_ n \longrightarrow V_{n + 1} \]
by the formula
14.27.0.1
\begin{equation} \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. \end{equation}
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*}
as desired.
Proof. We may choose a sequence $a = a_0, a_1, \ldots , a_ n = b$ of morphisms from $U$ to $V$ such that for each $i = 1, \ldots , n$ either there is a homotopy from $a_ i$ to $a_{i - 1}$ or there is a homotopy from $a_{i - 1}$ to $a_ i$. The calculation above shows that in this case either $s(a_ i)$ is homotopic to $s(a_{i - 1})$ as a map of chain complexes or $s(a_{i - 1})$ is homotopic to $s(a_ i)$ as a map of chain complexes. Of course, these things are equivalent and moreover being homotopic is an equivalence relation on the set of maps of chain complexes, see Homology, Section 12.13. This proves that $s(a)$ and $s(b)$ are homotopic as maps of chain complexes.
Next, we turn to $N(a)$ and $N(b)$. It 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 12.13.1. $\square$
Lemma 14.27.2. Let $\mathcal{A}$ be an additive category. Let $a : U \to V$ be a morphism of simplicial objects of $\mathcal{A}$. If $a$ is a homotopy equivalence, then $s(a) : s(U) \to s(V)$ is a homotopy equivalence of chain complexes. If in addition $\mathcal{A}$ is abelian, then also $N(a) : N(U) \to N(V)$ is a homotopy equivalence of chain complexes.
Proof. Omitted. See Lemma 14.27.1 above. $\square$
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