Corollary Let $f: X \rightarrow S$ be a Kan fibration between Kan complexes, let $s$ be a vertex of $S$, and set $X_{s} = \{ s\} \times _{S} X$. Then the image of the map $\pi _0(X_{s} ) \rightarrow \pi _0(X)$ is equal to the fiber of the map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(S)$ over the connected component $[s] \in \pi _0(S)$ determined by the vertex $s$. In other words, a vertex $x \in X$ satisfies $[ f(x) ] = [s]$ in $\pi _0(S)$ if and only if the connected component of $x$ has nonempty intersection with the fiber $X_{s}$.