10
$\begingroup$

Categories of manifolds (possibly with extra structure) tend not to have all colimits. Other questions have addressed when colimits of manifolds exist.

I'd like to know what we can say in general about those colimits which do happen to exist. In particular: given a colimit which does exist in some category $C$ of manifolds, I'd like to know whether it necessarily coincides with the underlying colimit in $\mathsf{Top}$ (and hence also $\mathsf{Set}$).

Hopefully, there is a general principle that I'm missing which answers this for many sensible categories of manifolds. But for the sake of a sharp question: does the forgetful functor $C \to \mathsf{Top}$ preserve colimits, for

  1. $C = \mathsf{Man}$, the category of topological manifolds and continuous functions [*]?
  2. $C = \mathsf{Diff}$, the category of smooth manifolds and smooth functions?
  3. $C = \mathsf{Rm}$, the category of smooth Riemannian manifolds and smooth local isometries?

Or maybe the functor $\mathsf{Man} \to \mathsf{Top}$ fails to preserve colimits, while forgetful $\mathsf{Diff} \to \mathsf{Man}$ or $\mathsf{Rm} \to \mathsf{Diff}$ does preserve colimits?

[*] Since being a topological manifold is a property of, rather than a structure on, a topological space, (1) isn't so much a forgetful functor as an inclusion.


I don't see any obvious right adjoints to these forgetful functors, although I'd be happy to learn otherwise.

A comment under this question suggests that there are examples of colimits which exist in $\mathsf{Man}$ (or maybe $\mathsf{Diff}$ in my notation?) but which don't agree with the underlying colimits in $\mathsf{Top}$. So a negative answer to (1) or (2) might spell out such an example.

I know also that there are various categories of 'generalized smooth spaces' (Frölicher spaces, diffeological spaces, Chen spaces, ...) with $\mathsf{Diff}$ as a subcategory. If the inclusion functor is known to preserve colimits, and colimits of the generalized smooth spaces are understood, then this could be useful to address (2). But I currently know nothing about such generalized smooth spaces, so a very helpful answer along these lines might indicate e.g. which versions of generalized smooth space are best able to answer (2), and whether any of these generalizations have been extended to the Riemannian case for (3).


Edit:

It seems that the category $\mathsf{Haus}$ of Hausdorff topological spaces and continuous maps is important to this story.

With this in mind, a better formulation of my question is as follows: we have a sequence of forgetful/inclusion functors $$ \mathsf{Rm} \rightarrow \mathsf{Diff} \rightarrow \mathsf{Man} \hookrightarrow \mathsf{Haus} \hookrightarrow \mathsf{Top}. $$ Which, if any, of these functors preserve all colimits?

The comment below by Martin Brandenburg already shows that the functors with target $\mathsf{Top}$ do not preserve all colimits.

$\endgroup$
6
  • 3
    $\begingroup$ (I'm a categorist not a topologist.) What you're asking about is called creating colimits. This is methodological advice not an answer: Coproducts no problem. Next think about filtered colimits of inclusions - what kind? probably open ones. Then quotients of equivalence relations - what kind? probably closed ones. Those are likely to be the colimits you want. There could be many more scary ones, so try out some topological counterexamples. $\endgroup$ Jun 28, 2021 at 16:14
  • 8
    $\begingroup$ $\mathbf{Man} \to \mathbf{Top}$ does not preserve epimorphisms (and hence, not all colimits), look at $\mathbb{R} \setminus \{0\} \hookrightarrow \mathbb{R}$. Probably you want to replace $\mathbf{Top}$ by $\mathbf{Haus}$, then at least epimorphisms are preserved. $\endgroup$ Jun 28, 2021 at 17:02
  • 3
    $\begingroup$ @PaulTaylor Creating colimits is much stronger, and it doesn't seem to be asked here. $\endgroup$ Jun 28, 2021 at 17:03
  • 1
    $\begingroup$ As @MartinBrandenburg says, the colimits are not "created" in this instance, because manifolds with the same topology could have different metrics, etc. I was just trying to give hints about how to ptove colimits or find counterexamples. $\endgroup$ Jun 29, 2021 at 16:30
  • 5
    $\begingroup$ Re. $\mathsf{Top}$ vs. $\mathsf{Haus}$: one motivation for sticking with $\mathsf{Top}$ is that we understand colimits there as gluings/quotients/disjoint unions just like in $\mathsf{Set}$. So a forgetful functor to $\mathsf{Top}$ failing to preserve colimits warns me that if I do stumble across a colimit of manifolds, it won't necessarily look like a gluing/quotient/disjoint union that happens to respect manifold/smooth/Riemannian structure. $\endgroup$ Jun 29, 2021 at 17:03

2 Answers 2

6
$\begingroup$

Here is an observation in the positive direction:

Observation: Let $\mathcal C$ be a cocomplete category, and assume that $I \in \mathcal C$ is a strong cogenerator -- that is, $Hom_{\mathcal C}(-,I) : \mathcal C^{op} \to Set$ is faithful and conservative. Let $\mathcal D \subseteq \mathcal C$ be a full subcategory which contains $I$. Then the inclusion $\iota: \mathcal D \to \mathcal C$ preserves colimits.

Proof: Suppose that $D = \varinjlim_j^{\mathcal D} D_j$ is a colimit in $\mathcal D$, and let $\underline D = \varinjlim_j^{\mathcal C} D_j$ be the colimit taken in $\mathcal C$. There is a canonical map $\varepsilon: \underline D \to D$. We have $Hom_{\mathcal C}(\underline D, I) = \varprojlim_j Hom_{\mathcal C}(D_j, I) = \varprojlim_j Hom_{\mathcal D}(D_j, I) = Hom_{\mathcal D}(D,I)$, so that $Hom(\varepsilon, I)$ is a bijection. Thus $\varepsilon$ is an isomorphism because $I$ is a strong cogenerator.

Example: Euclidean space $\mathbb R^d$ (for $d \geq 1$) is a strong [1] cogenerator in the category $Tych$ of Tychonoff spaces, which is contained in the full subcategory $Man \subset Tych$ of topological manifolds. Therefore any colimit of topological manifolds is a colimit of Tychonoff spaces.

[1] That it's a cogenerator follows almost by definition. So to see that $\mathbb R$ is a strong cogenerator in $Tych$, let $f : X \to Y$ be a map in $Tych$ which induces a bijection $f^\ast : C^0(Y) \to C^0(X)$; we wish to show that $f$ is a homeomorphism. First, $f$ is injective. For if $x \neq x' \in X$, then we may find a function $\varphi \in C^0(X)$ with $\varphi(x) \neq \varphi(x')$; since $\varphi$ extends along $f$ it follows that $f(x) \neq f(x')$. Next, $f$ is surjective. For if $y \in Y$ is not isolated and not in the image of $f$, then let $\psi \in C^0(Y)$ vanish uniquely at $y$; then $1/(f^\ast \psi) \in C^0(X)$ does not extend along $f$. If $f$ misses an isolated point $y \in Y$, then $f^\ast$ is not injective. Thus $f$ is a continuous bijection. If $f$ is not a homeomorphism, then pick a closed set $C \subseteq X$ whose image $f(C) \subseteq Y$ is not closed. Let $\varphi \in C^0(X)$ vanish uniquely on $C$. Then $\varphi$ does not extend continuously to $Y$.

$\endgroup$
4
  • $\begingroup$ This is cute. Upvote. $\endgroup$ Jul 8, 2021 at 14:41
  • $\begingroup$ We just need that $\hom(-,I)$ is conservative. $\endgroup$ Jul 8, 2021 at 23:52
  • 1
    $\begingroup$ True, but in a decent category this also implies faithfulness. A conservative functor preserving equalizers in a category with equalizers is also faithful. $\endgroup$ Jul 9, 2021 at 13:42
  • $\begingroup$ @IvanDiLiberti: There are plenty of “decent” categories without equalisers, though — including in particular most categories of manifolds. $\endgroup$ Jul 9, 2021 at 19:22
3
$\begingroup$

I'll write up Martin Brandenburg's comment as an answer:


Here is an example which shows that the inclusion functor $\mathsf{Man} \hookrightarrow \mathsf{Top}$ does not preserve colimits. Consider the span $\mathbb{R} \hookleftarrow \mathbb{R} \setminus \{ 0 \} \hookrightarrow \mathbb{R}$.

Taken in $\mathsf{Top}$, this span has pushout (whose nadir is) given by the 'line with two origins' obtained by gluing the two copies of $\mathbb{R}$ along $\mathbb{R} \setminus \{ 0 \}$.

Taken in $\mathsf{Man}$, the span also has a pushout, but here its nadir is just $\mathbb{R}$ itself: the inclusion $\mathbb{R} \setminus \{ 0 \} \hookrightarrow \mathbb{R}$ is an epimorphism in $\mathsf{Man}$. Any continuous maps $g, g' : \mathbb{R} \to X$ which coincide on $\mathbb{R} \setminus \{ 0 \}$ must also have the same value at $0$: by continuity, any open neighbourhoods of $g(0)$ and $g'(0)$ intersect on $g\left( (-\epsilon,0) \cup (0, \epsilon) \right) = g'\left( (-\epsilon,0) \cup (0, \epsilon) \right)$ for some sufficiently small $\epsilon > 0$. Since $X$ is a manifold, and so Hausdorff, this implies $g(0) = g'(0)$.

If we equip $\mathbb{R}$ with its usual smooth atlas, then all of the above carries through with $\mathsf{Diff}$ in place of $\mathsf{Man}$. And similarly, if we equip $\mathbb{R}$ also with its usual metric tensor, it carries through with $\mathsf{Rm}$ in place of $\mathsf{Diff}$. So, the example also works to show that forgetful functors $\mathsf{Diff} \to \mathsf{Top}$ and $\mathsf{Rm} \to \mathsf{Top}$ do not preserve colimits.

This does not tell us whether forgetting $\mathsf{Rm} \to \mathsf{Diff}$ or $\mathsf{Diff} \to \mathsf{Man}$ preserves colimits in general, though.


Edit:

As pointed out in comments, this example is really about Hausdorffness; in particular, it applies with $\mathsf{Haus}$ in place of $\mathsf{Man}$, so the inclusion $\mathsf{Haus} \hookrightarrow \mathsf{Top}$ does not preserve colimits.

$\endgroup$
2
  • 5
    $\begingroup$ I have a small objection: you are looking at the forgetful functor from Hausdorff manifolds to non-Haussdorff topological spaces and the problem directly comes from this Hausdorff vs non-Hausorff distinction. If we want to look at either "non-Hausdorff manifolds to Non-hausdorf spaces" or "Hausdorff manifolds to Hausdorff spaces" then this specific counter-example disappear. $\endgroup$ Jul 6, 2021 at 17:10
  • 1
    $\begingroup$ @Simon Henry: Yes, Hausdorffness is the key point here. I've edited the question and this answer to reflect this. Nonetheless, I think it's useful to have a counterexample illustrating that the forgetful functors to $\mathsf{Top}$ fail to preserve some colimits. $\endgroup$ Jul 8, 2021 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.