How to construct co-equalizers in the category $\mathbf{Top}$?

Well, do co-equalizers in $\mathbf{Top}$ exist at all?


If $f,g:X\to Y$ are two parallel morphisms, then the coequalizer $c:Y\to Z$ of $f,g$ will be the quotient map to $Z=Y/\sim$ where "$\sim$" is the smallest equivalence relation such that $f(x)\sim g(x)$. So it is the coequalizer in $\mathbf{Set}$ equipped with the quotient topology.

This is no coincidence. All colimits in $\mathbf{Top}$ have the same underlying set as the colimit of the underlying sets in $\mathbf{Set}$. In other words, the forgetful functor $U:\mathbf{Top}\to\mathbf{Set}$ preserves colimits. This is due to the fact that it has a right adjoint, namely the functor $\mathbf{Set}\to\mathbf{Top}$ which equips a set with the indiscrete topology.

Edit: Regarding the OP's comments, the adjointness only guarantees that IF the coequalizer exists, then its underlying set (and set map) is the coequalizer (diagram) in $\mathbf{Set}$.
To show that $Y/∼$ is the coequalizer in $\mathbf{Set}$, let $h:Y\to W$ be a function with $hf=hg$. Then the relation $h(y)=h(y')$ is an equivalence relation on $Y$ which contains the relation generated by $f(x)∼g(x)$ since $h(f(x))=h(g(x))$. Therefore the map $k:Y/∼\to W,\ k(c(y))=h(y)$ is well-defined and is obviously the only function $k:Y/∼\to W$ such that $kc=h$.
To show that $c$ is also the coequalizer in $\mathbf{Top}$, we just have to make sure that $c$ is continuous, and that the only possible function $k:Z\to W$ satisfying $kc=h$ is continuous. But $c$ is continuous because we equip $Z$ with the final topology, and this is also the reason why $k$ is continuous if $kc=h$ is. So $c:Y\to Z$ is the coequalizer in $\mathbf{Top}$.

Proving that $G:\mathbf{Set}\to\mathbf{Top},\ X\mapsto(X,\tau_{id})$ is right adjoint to $U$ is easy. Just note that $((X,\tau_{id}),\ 1:UGX\to X)$ is universal from $U$ to $X$: For each set map $f:Y\to X$ we have a continuous map $(Y,\tau_Y)\to(X,\tau_{id})$, namely $f$ itself.

  • $\begingroup$ But how to prove that it is really a coequalizer in Top? Can it be proved elementarily (without adjoints)? $\endgroup$ – porton Oct 26 '13 at 11:26
  • $\begingroup$ Also: How to prove that it has a right adjoint, namely the functor Set→Top which equips a set with the indiscrete topology? $\endgroup$ – porton Oct 26 '13 at 11:34
  • $\begingroup$ Adjointness warrants only that the limit in Top IF IT EXISTS coincides with the limit in Set, but not its existence. Right? $\endgroup$ – porton Oct 26 '13 at 11:53

Certainly they exist. If $F: J \to \mathbf{Top}$ is a small diagram, then the underlying set of $\text{colim}\; F$ is the colimit of $U \circ F$ as calculated in $\mathbf{Set}$, where $U: \mathbf{Top} \to \mathbf{Set}$ is the underlying-set functor. This set is equipped with the largest topology such that the functions $UF(j) \to \text{colim}\; UF$ specified by the universal cocone in $\mathbf{Set}$ are all continuous maps; this topological space is $\text{colim}\; F$.


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