The topological realization functor   reflect  coequalizers ? 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?
 A: 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.
