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Post Undeleted by Charles Rezk
2 Finished the answer.

Recall two facts about that geometric realization $T:S\to Top$ :

• (1) $T$ preserves all colimits.
• (2) $T$ preserves finite limits.

From 1.Thus, we see that it is enough to show that if $f:X\to Y$ is a map of simplicial sets such that $T(f)$ is a bijection, then $f$ is an isomorphism.

• (3) If $f:X\to Y$ is a map of simplicial sets such that, for every non-degenerate simplex $t$ of $Y_n$, there is a simplex $s$ of $X_n$ such that $f(s)=t$, then $f$ is surjective.
• (4) If

• Let $f:X\to Y$ is be a map of simplical sets such that (i) if $s\in X_n$ is non-degenerate, then $f(s)\in Y_n$ is non-degenerate, and (ii) if $s,s'\in X_n$ are non-degenerate and $f(s)=f(s')$, then $s=s'$, then s=s'$. It follows that$f$is injective. • If$f:X\to Y$is a map of simplicial sets such that for each non-degenerate$t\in Y_k$, there exists a non-degenerate$s$in some$X_n$such that$f(s)$is a degeneracy of$t$(or$f(s)=t$, when$n=k$), then$f$is surjective. • $$TX \approx \bigcup_n \bigcup_\sigma (\text{interior \Delta^n-\partial\Delta^n),$$where the$\sigma$range over non-degenerate$n$-simplices. It is not hard to see how this behaves as a functor: a map$f:X\to Y$induces a map$Tf:TX\to TY$which sends the boundaryless simplex of corresponding to a non-degenerate$\Delta^n$}).$$\sigma\in X_n$ to the bondaryless simplex corresponding to the non-degenerate $\tau\in Y_k$, where $\tau$ the non-degenerate simplex of which $f(\sigma)$ is degenerate (or, if $f(\sigma)$ is non-degnererate, then $f(\sigma)=\tau$.)The resulting map of boundaryless simplices $(\Delta^n-\partial \Delta^n)\to (\Delta^k-\partial \Delta^k)$ is described by a surjective map $[n]\to [k]$ in $\Delta$, and can be bijective only if $n=k$.

Given this, it's easy to check that if $Tf$ is a bijection, then the conditions of (1) and (2) must be satisfied, so $f$ must be bijective too.

Post Deleted by Charles Rezk
1