The Stacks project

Proof. Set $\mathcal{U}' = \mathcal{U} \times _\mathcal {X} \mathcal{X}$ equal to the fibre product as described in Categories, Lemma 4.32.3. Set $f' : \mathcal{U}' \to \mathcal{X}$ equal to the second projection. Then $\mathcal{U} \to \mathcal{U}'$, $u \mapsto (u, f(x), 1)$ is an equivalence over $\mathcal{X}$, hence we may replace $(\mathcal{U}, f)$ by $(\mathcal{U}', f')$ by Lemma 96.18.1. The advantage of this is that now $f'$ has a section $s'$ such that $f' \circ s' = \text{id}_\mathcal {X}$ on the nose. Namely, if $t : s \circ f \to \text{id}_\mathcal {X}$ is a $2$-isomorphism then we can set $s'(x) = (s(x), x, t_ x)$. Thus we may assume that $f \circ s = \text{id}_\mathcal {X}$.

In the case that $f \circ s = \text{id}_\mathcal {X}$ the result follows from general principles. We give the homotopy explicitly. Namely, for $n \geq 0$ define $s_ n : \mathcal{U}_ n \to \mathcal{U}_{n + 1}$ to be the $1$-morphism defined by the rule on objects

Define

as pullback along $s_ n$. We also set $s_{-1} = s$ and $h^0 : \Gamma (\mathcal{U}_0, f_0^{-1}\mathcal{F}) \to \Gamma (\mathcal{X}, \mathcal{F})$ equal to pullback along $s_{-1}$. Then the family of maps $\{ h^ n\} _{n \geq 0}$ is a homotopy between $1$ and $0$ on the extended Čech complex. $\square$