## 25.9 Homotopies

Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $L$ be a simplicial object of $\text{SR}(\mathcal{C}, X)$. According to Simplicial, Lemma 14.17.4 there exists an object $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], L)$ in the category $\text{Simp}(\text{SR}(\mathcal{C}, X))$ which represents the functor

\[ T \longmapsto \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\text{SR}(\mathcal{C}, X))}(\Delta [1] \times T, L) \]

There is a canonical morphism

\[ \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L) \to L \times L \]

coming from $e_ i : \Delta [0] \to \Delta [1]$ and the identification $\mathop{\mathrm{Hom}}\nolimits (\Delta [0], L) = L$.

Lemma 25.9.1. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $L$ be a simplicial object of $\text{SR}(\mathcal{C}, X)$. Let $n \geq 0$. Consider the commutative diagram

25.9.1.1
\begin{equation} \label{hypercovering-equation-diagram} \xymatrix{ \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L)_{n + 1} \ar[r] \ar[d] & (\text{cosk}_ n \text{sk}_ n \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L))_{n + 1} \ar[d] \\ (L \times L)_{n + 1} \ar[r] & (\text{cosk}_ n \text{sk}_ n (L \times L))_{n + 1} } \end{equation}

coming from the morphism defined above. We can identify the terms in this diagram as follows, where $\partial \Delta [n + 1] = i_{n!}\text{sk}_ n \Delta [n + 1]$ is the $n$-skeleton of the $(n + 1)$-simplex:

\begin{eqnarray*} \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L)_{n + 1} & = & \mathop{\mathrm{Hom}}\nolimits (\Delta [1] \times \Delta [n + 1], L)_0 \\ (\text{cosk}_ n \text{sk}_ n \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L))_{n + 1} & = & \mathop{\mathrm{Hom}}\nolimits (\Delta [1] \times \partial \Delta [n + 1], L)_0 \\ (L \times L)_{n + 1} & = & \mathop{\mathrm{Hom}}\nolimits ( (\Delta [n + 1] \amalg \Delta [n + 1], L)_0 \\ (\text{cosk}_ n \text{sk}_ n (L \times L))_{n + 1} & = & \mathop{\mathrm{Hom}}\nolimits ( \partial \Delta [n + 1] \amalg \partial \Delta [n + 1], L)_0 \end{eqnarray*}

and the morphism between these objects of $\text{SR}(\mathcal{C}, X)$ come from the commutative diagram of simplicial sets

25.9.1.2
\begin{equation} \label{hypercovering-equation-dual-diagram} \xymatrix{ \Delta [1] \times \Delta [n + 1] & \Delta [1] \times \partial \Delta [n + 1] \ar[l] \\ \Delta [n + 1] \amalg \Delta [n + 1] \ar[u] & \partial \Delta [n + 1] \amalg \partial \Delta [n + 1] \ar[l] \ar[u] } \end{equation}

Moreover the fibre product of the bottom arrow and the right arrow in (25.9.1.1) is equal to

\[ \mathop{\mathrm{Hom}}\nolimits (U, L)_0 \]

where $U \subset \Delta [1] \times \Delta [n + 1]$ is the smallest simplicial subset such that both $\Delta [n + 1] \amalg \Delta [n + 1]$ and $\Delta [1] \times \partial \Delta [n + 1]$ map into it.

**Proof.**
The first and third equalities are Simplicial, Lemma 14.17.4. The second and fourth follow from the cited lemma combined with Simplicial, Lemma 14.21.11. The last assertion follows from the fact that $U$ is the push-out of the bottom and right arrow of the diagram (25.9.1.2), via Simplicial, Lemma 14.17.5. To see that $U$ is equal to this push-out it suffices to see that the intersection of $\Delta [n + 1] \amalg \Delta [n + 1]$ and $\Delta [1] \times \partial \Delta [n + 1]$ in $\Delta [1] \times \Delta [n + 1]$ is equal to $\partial \Delta [n + 1] \amalg \partial \Delta [n + 1]$. This we leave to the reader.
$\square$

Lemma 25.9.2. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K, L$ be hypercoverings of $X$. Let $a, b : K \to L$ be morphisms of hypercoverings. There exists a morphism of hypercoverings $c : K' \to K$ such that $a \circ c$ is homotopic to $b \circ c$.

**Proof.**
Consider the following commutative diagram

\[ \xymatrix{ K' \ar@{=}[r]^-{def} \ar[rd]_ c & K \times _{(L \times L)} \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L) \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits (\Delta [1], L) \ar[d] \\ & K \ar[r]^{(a, b)} & L \times L } \]

By the functorial property of $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], L)$ the composition of the horizontal morphisms corresponds to a morphism $K' \times \Delta [1] \to L$ which defines a homotopy between $c \circ a$ and $c \circ b$. Thus if we can show that $K'$ is a hypercovering of $X$, then we obtain the lemma. To see this we will apply Lemma 25.7.1 to the pair of morphisms $K \to L \times L$ and $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], L) \to L \times L$. Condition (1) of Lemma 25.7.1 is satisfied. Condition (2) of Lemma 25.7.1 is true because $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], L)_0 = L_1$, and the morphism $(d^1_0, d^1_1) : L_1 \to L_0 \times L_0$ is a covering of $\text{SR}(\mathcal{C}, X)$ by our assumption that $L$ is a hypercovering. To prove condition (3) of Lemma 25.7.1 we use Lemma 25.9.1 above. According to this lemma the morphism $\gamma $ of condition (3) of Lemma 25.7.1 is the morphism

\[ \mathop{\mathrm{Hom}}\nolimits (\Delta [1] \times \Delta [n + 1], L)_0 \longrightarrow \mathop{\mathrm{Hom}}\nolimits (U, L)_0 \]

where $U \subset \Delta [1] \times \Delta [n + 1]$. According to Lemma 25.8.2 this is a covering and hence the claim has been proven.
$\square$

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