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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).