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Proposition Let $f: X \rightarrow S$ be a covering map of topological spaces. Then the induced map $\operatorname{Sing}_{\bullet }(f): \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(S)$ is a covering map of simplicial sets (in the sense of Definition

Proof. Let $\delta : X \rightarrow X \times _{S} X$ be the relative diagonal of $f$. We first claim $\delta $ exhibits $X$ as a summand of $X \times _{S} X$ in the category of topological spaces (that is, it is a homeomorphism of $X$ onto a closed and open subset of the fiber product $X \times _{S} X$). To verify this, we can work locally on $S$ and thereby reduce to the case where $X$ is a product of $S$ with a discrete topological space, in which case the result is clear. It follows that the induced map of singular simplicial sets

\[ \operatorname{Sing}_{\bullet }(\delta ): \operatorname{Sing}_{\bullet }(X) \hookrightarrow \operatorname{Sing}_{\bullet }(X \times _{S} X) \simeq \operatorname{Sing}_{\bullet }(X) \times _{ \operatorname{Sing}_{\bullet }(S)} \operatorname{Sing}_{\bullet }(X) \]

is also the inclusion of a summand (Remark, and is therefore a Kan fibration by virtue of Example Consequently, to show that $\operatorname{Sing}_{\bullet }(f)$ is a covering map, it will suffice to show that it is a Kan fibration (Remark This is a special case of Corollary, since $f: X \rightarrow S$ exhibits $X$ as a fiber bundle over $S$ (with discrete fibers). $\square$