Finitely cocomplete categories of compact Hausdorff spaces

Question

Edit: Zhen Lin incisively observes in a comment below that the category of compact Hausdorff spaces is monadic over the category of sets, hence is cocomplete. That answers the first part of question 1 as it was stated previously. I have changed that question accordingly.

Final edit: Chris Schommer-Pries answered the rest of the question in the negative with a very simple idea which I overlooked when thinking about this matter. Many thanks to Zhen Lin and Chris. To everyone else, I humbly apologize for asking such a simple question.

I am attempting to find finitely cocomplete categories of compact Hausdorff topological spaces with some convenient homotopical properties. In particular, I want to keep the nice feature of topological spaces that one can often perform homotopical constructions without using fibrant or cofibrant replacements (partly because all objects are fibrant). $\newcommand{\Top}{\mathrm{Top}} \newcommand{\CHTop}{\mathrm{CHTop}} \newcommand{\into}{\hookrightarrow}$

Let $\Top$ denote the category of topological spaces, and $\CHTop$ its full subcategory generated by the compact Hausdorff spaces. The first question is:

Question 1: Is the category $\CHTop$ finitely cocomplete (i.e. has all finite colimits)? Is every finite colimit of compact Hausdorff spaces in $\Top$ a Hausdorff space? In other words, does the inclusion $\CHTop\into\Top$ preserve finite colimits?

Regarding the plausibility of the question, it is certainly easy to construct quotients of compact Hausdorff spaces which are not Hausdorff. However, only certain types of quotients appear when computing finite colimits of spaces. I would still be inclined to believe the answer to the above question is no, yet I found no counterexample in my limited search. Also, I would be most interested to learn of simple conditions one can impose on spaces such that the corresponding full subcategory of $\CHTop$ is finitely cocomplete.

My second question is related to question 1. It concerns nice finitely cocomplete subcategories of compact Hausdorff spaces which still have "enough homotopies" to be able to avoid using fibrant/cofibrant replacements. As an example, simplicial sets do not have enough homotopies in the sense that one cannot in general concatenate homotopies since $\Delta^1\coprod_{\Delta^0} \Delta^1$ is not isomorphic to $\Delta^1$.

Question 2: Give examples of categories $C$ with a faithful functor $F:C\to\CHTop$ such that:

1. there exists an object $1_C$ in $C$ such that $F(1_C)$ is a singleton space;
2. there exists an object $I_C$ in $C$ such that $F(I_C)$ is homeomorphic to $I$;
3. $C$ admits all products with $I_C$, and the functor $F$ preserves those products;
4. $C$ is finitely cocomplete, and the composition $C\overset{F}{\to}\CHTop\into\Top$ preserves all finite colimits.

Bonus points if the category $C$ verifies the following strengthening of condition 3: $C$ has all finite products, and the functor $F$ preserves them.

Independently of the answer to question 1, I would be very interested to hear about any examples you know that would fit, or at least approximate, the requirements in question 2. I would be particularly interested in examples consisting of small, manageable spaces.

My own ill-determined idea to give an example as in question 2 is the subject of the following vague question.

Question 3: Does there exist a category as in question 2 which consists of finite polyhedron-like spaces, and where the morphisms are some sort of piecewise linear maps?

Answer 1

As Zhen Lin points out in the comments to your question, the category of compact Hausdorff spaces has all small colimits. However the inclusion functor from CHTop to Top does not preserve these colimits in general. It seems from your other questions that you are interested in a category, either CHTop or something similar, so that finite colimits exist and the inclusion functor preserves finite colimits.

This will not be possible. In particular the answer to question 2 (hence also question 3) is no, as the following example illustrates.

Let $X = Y = I = [0,1]$ and consider a coequalizer $Y \rightrightarrows X$ where the first map is the identity and the second map is: $$ t \mapsto 2(t - t^2) $$ Even though Y is compact, this generates an equivalence relation with non-closed equivalence classes. The equivalences classes are essentially the solution sets to the above non-linear recurrence relation. In particular nearly every equivalence class cannot be separated from the middle fixed point $t = 0.5$, and hence the quotient is non-Hausdorff.

Answer 2

The examples given are part of the case for homotopy colimits. For example the pushout of the two maps $$S^1 \leftarrow S^1 \to S^!$$ given by $z\mapsto z^2, z \mapsto z^3$ is not Hausdorff. But the double mapping cylinder $M$ is a nice CW-complex. Amusingly, this is rel;ated to the case for groupoids, where the pushout in groups of the two maps $$ \mathbb Z \leftarrow \mathbb Z \to \mathbb Z$$ is the trefoil group $T$ with generators $x,y$ and relation $x^2=y^3$, but the homotopy pushout in groupoids is the "trefoil groupoid", say $T'$, with two objects $0,1$, generators $x,y$ at $0,1$ respectively and one arrow $\iota :0 \to 1$ conjugating $x^2$ to $y^3$. The advantage of this is that it "separates" the two group generators $x,y$, and of course $T'$ is the fundamental groupoid of $M$ on two base points, one at each end of the cylinder.