Suppose $f,g\colon X \to Y$ are maps of compact Hausdorff spaces $X$ and $Y$. Their coequalizer $\pi \colon Y \to Z$ is the set $Z:=Y/\sim$, where $\sim$ is the equivalence relation generated by $f(x)\sim g(x)$. The map $\pi$ sends $y$ to its equivalence class, and $Z$ is given the quotient topology. It’s easy to see that $Z$ is compact, but it may not be Hausdorff. However, under the following kind of strange assumption, one can show that $Z$ is always Hausdorff. Assume that there are functions $\iota\colon Z \to Y$ and $s\colon Y \to X$—these are not assumed to be continuous—such that $$ \pi \iota = 1_Z,\quad gs = 1_Y,\quad fs = \iota\pi. $$
In this situation, the foregoing data is altogether called a “$U$-split coequalizer” for reasons which I’ll say a bit about at the end.
I was trying to come up with an interesting example, e.g. one where the maps $s$ and $\iota$ are not continuous, and I failed. I am looking for an example of the above situation where $s$ and/or $\iota$ is not continuous, or at least one that isn’t just collapsing pieces of a disjoint union.
Here is a non-interesting example: Let $$X=[0,1]\times\{0,1,2\},\quad Y = [0,1]\times \{0,1\}.$$ Define $f$ and $g$ as follows. $$f(t,n) = \begin{cases} (t,0) & n = 0,1 \\ (t,1) & n = 2 \end{cases}\qquad g(t,n) = \begin{cases} (t,0) & n = 1 \\ (t,1) & n = 1,2\end{cases}$$
It’s not hard to see that $Z = [0,1]$, that $\iota\colon Z \to Y$ is perhaps $t\mapsto (t,0)$, and that $s\colon Y \to X$ could be $s(t,n) = (t,n)$.
Here’s a short explanation of the category theory mumbo jumbo motivating this question: Let $U\colon$ cHaus $\to$Set be the forgetful functor from the category of compact Hausdorff spaces and continuous functions to the category of sets and functions. As part of Beck’s monadicity theorem, to prove that cHaus is monadic over Set, one needs to show that $U$ creates coequalizers of $U$-split parallel pairs.
That is, given just the data of $f,g\colon X\to Y$, the following is true. If on the level of underlying sets, we can find $\pi$, $\iota$, $s$ and $Z$ exactly as above, only missing the assumption that $Z$ is a topological space and $\pi$ is continuous, then actually $Z$ can be given a topology that makes $\pi$ continuous and satisfy the universal property of the coequalizer, which I’d prefer not to define. In other words, given a coequalizer $\pi\colon Y \to Z$ of underlying sets with the extra information of the “splitting” functions $\iota$ and $s$, then actually $\pi\colon Y\to Z$ is the underlying function of sets of a continuous function. We say $U$ creates this coequalizer.