Does the category of topological spaces with an action of a topological monoid admit coequalizers? Let $\mathsf{Top}$ be the category of topological spaces with continuous maps and let $T$ be a monoid in $\mathsf{Top}$. Let $f,g: M \to N$ be two parallel maps in $\mathsf{Top}_T$, the category of topological spaces with a continuous action of $T$.
I would like to know whether there exists the coequalizer of $(f,g)$ in $\mathsf{Top}_T$.
Let me sketch what I know.
The forgetful functor $U : \mathsf{Top}_T \to \mathsf{Set}_T$ has a right adjoint  (endow $T$-sets with the indiscrete topology), hence $U$ preserves colimits. In particular, if $\mathsf{Coeq}(f,g)$ exists in $\mathsf{Top}_T$, then it has to be the quotient of $N$ by the smallest equivalence relation containing $\{ (f(x),g(x)) \mid x \in X \}$ (that is, the coequalizer in $\mathsf{Set}_T$) endowed with a suitable topology.
Denote it by $Q$, with projection $q : N \to Q$.
If $T$ is core-compact, I know that $T \times - : \mathsf{Top} \to \mathsf{Top}$ preserves colimits (because it admits a right adjoint) and hence $(Q,q)$ with the quotient topology on $Q$ would be the coequalizer.
If $T$ is arbitrary, I thought I may consider the finest topology on $Q$ which also makes the $T$-action $T \times Q \to Q$ a continuous map (so, the finest in the quotient topology such that $T \times Q \to Q$ is continuous). However, I am stuck in proving that $(Q,q)$ with the latter topology on $Q$ is universal in $\mathsf{Top}_T$.
In particular, if $(C,c)$ is any test object in $\mathsf{Top}_T$ with $c \circ f = c \circ g$, then I am not able to prove that the unique map $\tilde{c} : Q \to C$ such that $\tilde{c} \circ q = c$, provided by the universal property of $(Q,q)$ in $\mathsf{Set}_T$, is continuous.
Does anybody have a suggestion?
Or does there exist a counter-example which shows that, in general, such coequalizer does not exist?
 A: Rather than trying to construct some specific concrete $T$-space that will be the coequalizer, you can instead just build a thing that manifestly has the universal property by cobbling together every possible candidate.  For instance, let $(C_i,c_i)_{i\in I}$ be a family of representatives of all the isomorphism classes of a $T$-space $C$ and a surjective morphism $c:N\to C$ that coequalizes $f$ and $g$.  The $c_i$ then combine to give a map $h:N\to\prod_i C_i$.  The coequalizer of $f$ and $g$ is then just the image of this map $h$, as a sub-$T$-space of the product $\prod_i C_i$.  The universal property is trivial to verify since every instance of it is witnessed by just taking the projection onto one of the factors in the product.
From a broader perspective, this construction is very similar to the proof of the general adjoint functor theorem, and indeed can be considered as a special case of it.  We are constructing the coequalizer by taking the limit of all possible candidate coequalizers, restricting ourselves to those where $c:N\to C$ is surjective to make sure that this limit is essentially small so we can be sure it exists.  More generally, if you have a category with all limits, then it is very nearly guaranteed to have all colimits as well, since you can construct the colimit of a diagram by taking the limit of all the cocones.  The only issue with this is that this is a priori a large limit that might not exist; this is where the "solution set" condition of the adjoint functor theorem comes in to reduce it to a small limit.  In practice, this solution set condition is almost always trivial to verify, and just amounts to saying that every cocone factors through one of bounded cardinality.
