The topological realization functor reflect coequalizers ?

2010-11-15T22:15:33Z

Reading the book of Goess-Jardine or of Gabriel-Zisman on the simplicial homotopy there are coker presentation of the boundary $\partial\Delta^n $ of the elementary simplex $\Delta^n $ or the horn $\Lambda^n_k $ (see GOerss-Jardine "Simplicial Homotopy Theory" pag.9) .

Now the image of this presentation coker diagram by the topological realization functor $T: S \to Top$ (where $S$ category of functors from $\Delta^{ op }$ to $Set$, and $Top$ category of topological spaces) is of course still a coker diagram (the functor $T$ preserves colimits being a left adjoint) , a separate verification that this latter is a coequalizers (of topological spaces) is evident (or at most geometrically intuitive). Of course is a no hard problem give a demonstration that also the original presentation (by simplicial spaces) is a coequalizer.

Anyway this little question has suggested me the follow question:

Let $\iota: \coprod \Delta \subset S$ the subcategory by object the finite coproducts of representales , and by morphisms constructed from morphisms between representable and coprojections. Considering the restriction functor $T\circ \iota: \coprod \Delta \to Top $, the question is:

This functors reflex cokernels?

Answer by Charles Rezk for The topological realization functor reflect coequalizers ?

2010-12-01T21:50:20Z

I claim that geometric realization reflects colimits of all sorts. That is, given a functor $F: C\to sSet$ and a cocone ${F(C_i)\to X}$ in $sSet$ such that $\newcommand{\colim}{\operatorname{colim}}T(\colim_C F)\to T(X)$ is a bijection, then $\colim_C F\to X$ is an isomorphism of simplicial sets.

Recall that geometric realization $T:S\to Top$ preserves all colimits. Thus, 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.

To show this, I want to use the notion of a nongenerate simplex of a simplicial set $X$, and the fact that every simplex $x\in X_n$ is the degeneracy of a unique non-degenerate simplex $t\in X_k$, and is so in a unique way (i.e., there is a unique surjective map $\sigma:[n]\to [k]$ in $\Delta$ such that $(X\sigma)(t)=x$; this fact is sometimes called the "Eilenberg-Zilber lemma".)

Given this, it is not hard to show the following.

Let $f:X\to Y$ 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 $s=s'$. 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.

Now consider the geometric realization $TX$ of a simplicial set $X$. As a set, this has the form of a disjoint union $$ TX \approx \bigcup_n \bigcup_\sigma (\Delta^n-\partial\Delta^n),$$ 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 $\sigma\in X_n$ 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.

The topological realization functor reflect coequalizers ? - MathOverflow

The topological realization functor reflect coequalizers ?

Answer by Charles Rezk for The topological realization functor reflect coequalizers ?