Why is the Cech nerve $C(U)$ of a surjective map $U\to X$ weakly equivalent to $X$?

Question

Let $f:U\to X$ be a surjective map of sets and $$ ...U\times_XU\times_XU \substack{\textstyle\rightarrow\\[-0.6ex] \textstyle\rightarrow \\[-0.6ex] \textstyle\rightarrow} U\times_X\times U \substack{\textstyle\rightarrow\\[-0.6ex] \\[-0.6ex] \textstyle\rightarrow} U $$ the simplicial set given by its Cech nerve $C(U)$. When I consider $X$ as a discrete simplicial set, there is a map $f':C(U)\to X$ of simplicial sets.

Why is $f'$ a weak equivalence?

I see that $colim(U\times_X\times U \substack{\textstyle\rightarrow\\[-0.6ex] \\[-0.6ex] \textstyle\rightarrow} U)\cong X$ and that $\pi_0(C(U))$ is that colimit but I don't understand why there is a weak equivalence between the simplicial set $C(U)$ and the discrete one $\pi_0(C(U))$? Is $C(U)$ itself discrete? I don't think so.

Update: Is $C(U)$ weakly equivalent to $\pi_0(C(U))$ even for a non-surjective map $f:U\to X$ of sets?

Answer 1

Using the axiom of choice, $f : U \to X$ is a split epimorphism, and so the Čech nerve has an extra degeneracy. As usual, this extra degeneracy can be used to construct a homotopy equivalence between the Čech nerve of $f$ and $X$; and of course, every homotopy equivalence is in particular a weak homotopy equivalence.

Answer 2

Show that $f': C(U)\to X$ is an acyclic fibration. Using the fact that $C(U)$ is a groupoid, it suffices to verify two conditions.