Lemma 14.29.3. Let $\mathcal{A}$ be an abelian category. Let $U$, $V$ be simplicial objects of $\mathcal{A}$. Let $a, b : U \to V$ be a pair of morphisms. Assume the corresponding maps of chain complexes $N(a), N(b) : N(U) \to N(V)$ are homotopic by a homotopy $\{ N_ n : N(U)_ n \to N(V)_{n + 1}\} $. Then there exists a homotopy from $a$ to $b$ as in Definition 14.26.1. Moreover, one can choose the homotopy $h : U \times \Delta [1] \to V$ such that $N_ n = N(h)_ n$ where $N(h)$ is the homotopy coming from $h$ as in Section 14.27.
Proof. Let $(\diamond N(U), \diamond a, \diamond b, \diamond h)$ be as in Lemma 14.29.1 and its proof. By that lemma there exists a morphism $\diamond N(U) \to N(V)$ representing the triple $(N(a), N(b), \{ N_ n\} )$. We will show there exists a morphism $\psi : N(U \times \Delta [1]) \to \diamond {N(U)}$ such that $\diamond a = \psi \circ N(e_0)$, and $\diamond b = \psi \circ N(e_1)$. Moreover, we will show that the homotopy between $N(e_0), N(e_1) : N(U) \to N(U \times \Delta [1])$ coming from (14.27.0.1) and Lemma 14.27.1 with $h = \text{id}_{U \times \Delta [1]}$ is mapped via $\psi $ to the canonical homotopy $\diamond h$ between the two maps $\diamond a, \diamond b : N(U) \to \diamond {N(U)}$. Certainly this will imply the lemma.
Note that $N : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})$ as a functor is a direct summand of the functor $s : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})$. Also, the functor $\diamond $ is compatible with direct sums. Thus it suffices instead to construct a morphism $\Psi : s(U \times \Delta [1]) \to \diamond {s(U)}$ with the corresponding properties. This is what we do below.
By Definition 14.26.1 the morphisms $e_0 : U \to U \times \Delta [1]$ and $e_1 : U \to U \times \Delta [1]$ are homotopic with homotopy $\text{id}_{U \times \Delta [1]}$. By Lemma 14.27.1 we get an explicit homotopy $\{ h_ n : s(U)_ n \to s(U \times \Delta [1])_{n + 1}\} $ between the morphisms of chain complexes $s(e_0) : s(U) \to s(U \times \Delta [1])$ and $s(e_1) : s(U) \to s(U \times \Delta [1])$. By Lemma 14.29.2 above we get a corresponding morphism
\[ \Phi : \diamond {s(U)} \to s(U \times \Delta [1]) \]
According to the construction, $\Phi _ n$ restricted to the summand $s(U)[-1]_ n = s(U)_{n - 1}$ of $\diamond {s(U)}_ n$ is equal to $h_{n - 1}$. And
\[ h_{n - 1} = \sum \nolimits _{i = 0}^{n - 1} (-1)^{i + 1} s^ n_ i \cdot \alpha ^ n_{i + 1} : U_{n - 1} \to \bigoplus \nolimits _ j U_ n \cdot \alpha ^ n_ j. \]
with obvious notation.
On the other hand, the morphisms $e_ i : U \to U \times \Delta [1]$ induce a morphism $(e_0, e_1) : U \oplus U \to U \times \Delta [1]$. Denote $W$ the cokernel. Note that, if we write $(U \times \Delta [1])_ n = \bigoplus _{\alpha : [n] \to [1]} U_ n \cdot \alpha $, then we may identify $W_ n = \bigoplus _{i = 1}^ n U_ n \cdot \alpha ^ n_ i$ with $\alpha ^ n_ i$ as in Section 14.26. We have a commutative diagram
\[ \xymatrix{ 0 \ar[r] & U \oplus U \ar[rd]_{(1, 1)} \ar[r] & U \times \Delta [1] \ar[d]^\pi \ar[r] & W \ar[r] & 0 \\ & & U & & } \]
This implies we have a similar commutative diagram after applying the functor $s$. Next, we choose the splittings $\sigma _ n : s(W)_ n \to s(U \times \Delta [1])_ n$ by mapping the summand $U_ n \cdot \alpha ^ n_ i \subset W_ n$ via $(-1, 1)$ to the summands $U_ n \cdot \alpha ^ n_0 \oplus U_ n \cdot \alpha ^ n_ i \subset (U \times \Delta [1])_ n$. Note that $s(\pi )_ n \circ \sigma _ n = 0$. It follows that $(1, 1) \circ \delta (\sigma )_ n = 0$. Hence $\delta (\sigma )$ factors as in Lemma 14.29.2. By that lemma we obtain a canonical morphism $\Psi : s(U \times \Delta [1]) \to \diamond {s(U)}$.
To compute $\Psi $ we first compute the morphism $\delta (\sigma ) : s(W) \to s(U)[-1] \oplus s(U)[-1]$. According to Homology, Lemma 12.14.4 and its proof, to do this we have compute
\[ d_{s(U \times \delta [1]), n} \circ \sigma _ n - \sigma _{n - 1} \circ d_{s(W), n} \]
and write it as a morphism into $U_{n - 1} \cdot \alpha ^{n - 1}_0 \oplus U_{n - 1} \cdot \alpha ^{n - 1}_ n$. We only do this in case $\mathcal{A}$ is the category of abelian groups. We use the short hand notation $x_{\alpha }$ for $x \in U_ n$ to denote the element $x$ in the summand $U_ n \cdot \alpha $ of $(U \times \Delta [1])_ n$. Recall that
\[ d_{s(U \times \delta [1]), n} = \sum \nolimits _{i = 0}^ n (-1)^ i d^ n_ i \]
where $d^ n_ i$ maps the summand $U_ n \cdot \alpha $ to the summand $U_{n - 1} \cdot (\alpha \circ \delta ^ n_ i)$ via the morphism $d^ n_ i$ of the simplicial object $U$. In terms of the notation above this means
\[ d_{s(U \times \delta [1]), n}(x_\alpha ) = \sum \nolimits _{i = 0}^ n (-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} \]
Starting with $x_\alpha \in W_ n$, in other words $\alpha = \alpha ^ n_ j$ for some $j \in \{ 1, \ldots , n\} $, we see that $\sigma _ n(x_\alpha ) = x_\alpha - x_{\alpha ^ n_0}$ and hence
\[ (d_{s(U \times \delta [1]), n} \circ \sigma _ n)(x_\alpha ) = \sum \nolimits _{i = 0}^ n (-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} - \sum \nolimits _{i = 0}^ n (-1)^ i (d^ n_ i(x))_{\alpha ^ n_0 \circ \delta ^ n_ i} \]
To compute $d_{s(W), n}(x_\alpha )$, we have to omit all terms where $\alpha \circ \delta ^ n_ i = \alpha ^{n - 1}_0, \alpha ^{n - 1}_ n$. Hence we get
\[ \begin{matrix} (\sigma _{n - 1} \circ d_{s(W), n})(x_\alpha ) =
\\ \sum \nolimits _{i = 0, \ldots , n\text{ and } \alpha \circ \delta ^ n_ i \not= \alpha ^{n - 1}_0\text{ or }\alpha ^{n - 1}_ n} \Big((-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} - (-1)^ i (d^ n_ i(x))_{\alpha ^{n - 1}_0} \Big)
\end{matrix} \]
Clearly the difference of the two terms is the sum
\[ \sum \nolimits _{i = 0, \ldots , n\text{ and } \alpha \circ \delta ^ n_ i = \alpha ^{n - 1}_0\text{ or }\alpha ^{n - 1}_ n} \Big((-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} - (-1)^ i (d^ n_ i(x))_{\alpha ^{n - 1}_0} \Big) \]
Of course, if $\alpha \circ \delta ^ n_ i = \alpha ^{n - 1}_0$ then the term drops out. Recall that $\alpha = \alpha ^ n_ j$ for some $j \in \{ 1, \ldots , n\} $. The only way $\alpha ^ n_ j \circ \delta ^ n_ i = \alpha ^{n - 1}_ n$ is if $j = n$ and $i = n$. Thus we actually get $0$ unless $j = n$ and in that case we get $(-1)^ n(d^ n_ n(x))_{\alpha ^{n - 1}_ n} - (-1)^ n(d^ n_ n(x))_{\alpha ^{n - 1}_0}$. In other words, we conclude the morphism
\[ \delta (\sigma )_ n : W_ n \to (s(U)[-1] \oplus s(U)[-1])_ n = U_{n - 1} \oplus U_{n - 1} \]
is zero on all summands except $U_ n \cdot \alpha ^ n_ n$ and on that summand it is equal to $((-1)^ nd^ n_ n, -(-1)^ nd^ n_ n)$. (Namely, the first summand of the two corresponds to the factor with $\alpha ^{n - 1}_ n$ because that is the map $[n - 1] \to [1]$ which maps everybody to $0$, and hence corresponds to $e_0$.)
We obtain a canonical diagram
\[ \xymatrix{ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & \diamond {s(U)} \ar[r] \ar[d]^{\Phi }& s(U)[-1] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & s(U \times \Delta [1]) \ar[r] \ar[d]^\Psi & s(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] & \diamond {s(U)} \ar[r] & s(U)[-1] \ar[r] & 0 } \]
We claim that $\Phi \circ \Psi $ is the identity. To see this it is enough to prove that the composition of $\Phi $ and $\delta (\sigma )$ as a map $s(U)[-1] \to s(W) \to s(U)[-1] \oplus s(U)[-1]$ is the identity in the first factor and minus identity in the second. By the computations above it is $((-1)^ nd^ n_0, -(-1)^ nd^ n_0) \circ (-1)^ n s^ n_ n = (1, -1)$ as desired. $\square$
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