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On Kerodon (subsection 1.1.8) the following definition is given for the geometric realization of a simplicial set:

Let $S_{\bullet}$ be a simplicial set and let $Y$ be a topological space. We will say that a map of simplicial sets $u: S_{\bullet} \to \operatorname{Sing_{\bullet}}(Y)$ exhibits $Y$ as a geometric realization of $S_{\bullet}$ if, for every topological space $X$, the composite map $$ \operatorname{Hom_{Top}}(Y, X) \rightarrow \operatorname{Hom_{Set_{\Delta }}}( \operatorname{Sing_{\bullet}}( Y ), \operatorname{Sing_{\bullet}}(X)) \xrightarrow{ \circ u} \operatorname{Hom_{Set_{\Delta }}}( S_{\bullet }, \operatorname{Sing_{\bullet}}(X) ) $$ is bijective.

Then two claims are made:

  • If $u$ exhibits $Y$ as a geometric realization of $S_{\bullet}$, then $Y$ is determined up to homeomorphism;
  • The space $Y$ depends functorially on $S_{\bullet}$.

I want to show functoriality - the second claim then follows since functors preserve isos. So assume $u$ exhibits $Y$ as a geometric realization of $S_{\bullet}$ and $v$ exhibits $Y'$ as a geometric realization of $T_{\bullet}$, and let $f: S_{\bullet} \to T_{\bullet}$ be a map of simplicial sets. Using bijectivity of the upper horizontal composition in this diagram

Diagram

I can, for any fixed $X$, obtain a unique map $\operatorname{Hom_{Top}}(Y', X) \to \operatorname{Hom_{Top}}(Y, X)$ such that the outer square commutes. But I don't see how I can use this observation to obtain a map $F(f): Y \to Y'$ which induces this map and makes all squares commmute. Can anyone complete the argument (if what I've done so far is correct)? The claims are supposed to follow 'immediately from the definitions', so it's quite likely I'm overlooking something obvious.

Thanks in advance, and please let me know if you'd like me to clarify anything.

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1 Answer 1

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You have constructed a natural transformation $\mathrm{Hom}_{\mathrm{Top}}(Y', -) \to \mathrm{Hom}_{\mathrm{Top}}(Y, -)$. By Yoneda, this is the same thing as a map $Y \to Y'$ in $\mathrm{Top}$.

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