The Stacks project

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

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

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

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

and

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

Proof. Denote $(p_ n, q_ n) : B_ n \to A_ n \oplus A_ n$ the morphism $\pi _ n$ of Homology, Lemma 12.14.4. Also write $(a, b) : A \oplus A \to B$, and $r : B \to C$ for the maps in the short exact sequence. Write the factorization of $\delta (s)$ as $\delta (s) = (1, -1) \circ f$. This means that $p_{n - 1} \circ d_{B, n} \circ s_ n = f_ n$, and $q_{n - 1} \circ d_{B, n} \circ s_ n = - f_ n$, and Set $B_ n \to \diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$ equal to $(p_ n, q_ n, f_ n \circ r_ n)$.

Now we have to check that this actually defines a morphism of complexes. We will only do this in the case of abelian groups. Pick $x \in B_ n$. Then $x = a_ n(x_1) + b_ n(x_2) + s_ n(x_3)$ and it suffices to show that our definition commutes with differential for each term separately. For the term $a_ n(x_1)$ we have $(p_ n, q_ n, f_ n \circ r_ n)(a_ n(x_1)) = (x_1, 0, 0)$ and the result is obvious. Similarly for the term $b_ n(x_2)$. For the term $s_ n(x_3)$ we have

by definition of $f_ n$. And

The result follows as $f$ is a morphism of complexes. $\square$

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

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

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

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

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

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

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

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

Clearly the difference of the two terms is the sum

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

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

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. $\square$