Lemma 14.26.9. Let $f : Y \to X$ be a morphism of a category $\mathcal{C}$ with fibre products. Assume $f$ has a section $s$. Consider the simplicial object $U$ constructed in Example 14.3.5 starting with $f$. The morphism $U \to U$ which in each degree is the self map $(s \circ f)^{n + 1}$ of $Y \times _ X \ldots \times _ X Y$ given by $s \circ f$ on each factor is homotopic to the identity on $U$. In particular, $U$ is homotopy equivalent to the constant simplicial object $X$.

Proof. Set $g^0 = \text{id}_ Y$ and $g^1 = s \circ f$. We use the morphisms

\begin{eqnarray*} Y \times _ X \ldots \times _ X Y \times \mathop{\mathrm{Mor}}\nolimits ([n], ) & \to & Y \times _ X \ldots \times _ X Y \\ (y_0, \ldots , y_ n) \times \alpha & \mapsto & (g^{\alpha (0)}(y_0), \ldots , g^{\alpha (n)}(y_ n)) \end{eqnarray*}

where we use the functor of points point of view to define the maps. Another way to say this is to say that $h_{n, 0} = \text{id}$, $h_{n, n + 1} = (s \circ f)^{n + 1}$ and $h_{n, i} = \text{id}_ Y^{i + 1} \times (s \circ f)^{n + 1 - i}$. We leave it to the reader to show that these satisfy the relations of Lemma 14.26.2. Hence they define the desired homotopy. See also Remark 14.26.4 which shows that we do not need to assume anything else on the category $\mathcal{C}$. $\square$

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