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Corollary Let $X$ and $Y$ be simplicial sets. Then the canonical map $\theta _{X,Y}: |X \times Y| \rightarrow |X| \times |Y|$ is a bijection. If either $X$ or $Y$ is finite, then $\theta $ is a homeomorphism.

Proof. The first assertion follows immediately from Theorem If $X$ and $Y$ are both finite, then the product $X \times Y$ is also finite (Remark, so that the geometric realizations $|X|$, $|Y|$, and $|X \times Y|$ are compact Hausdorff spaces (Corollary In this case, $\theta _{X,Y}$ is a continuous bijection between compact Hausdorff spaces, and therefore a homeomorphism.

Now suppose that $X$ is finite and $Y$ is arbitrary. Let $M = \operatorname{Hom}_{\operatorname{Top}}( |X|, |X \times Y| )$ denote the set of all continuous functions from $|X|$ to $|X \times Y|$, endowed with the compact-open topology. For every finite simplicial subset $Y' \subseteq Y$, the composite map

\[ |X| \times |Y'| \xrightarrow { \theta _{X,Y'}^{-1} } |X \times Y'| \hookrightarrow |X \times Y|, \]

determines a continuous function $\rho _{Y'}: |Y'| \rightarrow M$. Writing the geometric realization $|Y|$ as a colimit $\varinjlim _{Y' \subseteq Y} |Y'|$ (see Remark, we can amalgamate the functions $f_{Y'}$ to a single continuous function $\rho : |Y| \rightarrow M$. Our assumption that $X$ guarantees that the topological space $|X|$ is compact and Hausdorff, so the evaluation map

\[ \operatorname{ev}: |X| \times M \rightarrow |X \times Y| \quad \quad (x,f) \mapsto f(x) \]

is continuous (see Theorem ). We complete the proof by observing that the bijection $\theta _{X,Y}^{-1}$ is a composition of continuous functions

\[ |X| \times |Y| \xrightarrow { \operatorname{id}\times \rho } |X| \times M \xrightarrow {\operatorname{ev}} | X \times Y |, \]

and is therefore continuous. $\square$