This question tells us that in general colimits do not exist in the category of manifolds.

However, this negative answer is not very satisfying. A manifold can be considered as a colimit of its altas. In this sense, it seems that constructing manifolds via atlas is a means of completing the model category of open subsets of $R^n$ (or more precisely a comma category based on this), so it should be reasonable that at least certain kinds of colimits should hold in the category of manifolds.

Considering the altas as a "good" system, do colimits of systems which are
1) of same-dimensional manifolds
2) have only open injections as morphisms,
3) is countable
4) (add your additional condition here),

PS the counter example in the above referred question fails conditions 1 and 2.

EDIT: After our good discussion below, perhaps let me restate my original question. I am motivated by the fact that a manifold is a colimit of its altas. What I'm looking for is a generalization of this fact. Can we extract out some general properties of an atlas that makes its colimit exist?

For instance do colimits of countable inverse limits of open smooth injections of (neccessarily) same-dimensional manifolds exist? (remove as many quantifiers as is uncessary).

I really do mean inverse system and not direct system. Observe that the altas (if we take all finite intersections) is filtered in the direction away from the colimit.

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    $\begingroup$ Countable coproducts exist, and it seems plausible to me that (countable) filtered colimits of open embeddings would exist. I would imagine coequalizers are problematic. $\endgroup$ – Todd Trimble Sep 13 '10 at 13:41
  • $\begingroup$ Since you mention below that you are interested in the analytic category, let me add that there is a very general notion of manifold based on the notion of pseudogroup of homeomorphisms, which covers many special cases: topological, smooth, analytic, complex, foliated, elliptic, hyperbolic, etc. (reference: Thurston's Three-Dimensional Geometry and Topology). It seems to me a really good answer to your question ought to work in this generality. $\endgroup$ – Todd Trimble Sep 13 '10 at 16:07
  • $\begingroup$ on your earlier comment, Todd, flitered colimits do not cover the case of an atlas. Atlases are filtered in the other direction. It seems queer that the 'obvious' colimits which exist don't include the colimit we use to construct manifolds. $\endgroup$ – user2529 Sep 14 '10 at 5:02

This is a few years late, but here goes.

Associated to an open cover $\{U_i\}$ of a (Hausdorff paracompact) manifold $M$ there is a cover groupoid, whose space of objects is the disjoint union ${\mathcal U}:= \sqcup U_i$ and whose space of arrows is the fiber product ${\mathcal U }\times _M {\mathcal U}$. The manifold $M$ is the quotient space of the cover groupoid, i.e., the colimit of ${\mathcal U }\times _M {\mathcal U}\rightrightarrows {\mathcal U}$. Note that the cover groupoid is proper so its quotient is Hausdorff. The orbit spaces of two Morita equivalent groupoids are isomorphic. Hence you could write a manifold as a colimit of a Lie groupoid which is Morita equivalent to a cover groupoid. I believe the conditions for being equivalent to a cover groupoid are being proper and having all isotropy groups trivial ( i.e., $Hom (x,x)$ is a trivial group for any object $x$ of your Lie groupoid).

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    $\begingroup$ Eugene, can you axiomatize a "cover groupoid" intrinsically among groupoid objects in ${\mathsf{Mfd}}$? In other words, I hope to start with a groupoid object $U$ in ${\mathsf{Mfd}}$ and construct a manifold as the colimit of $U\times_{U_0} U \rightrightarrows U$. $\endgroup$ – user2529 Nov 13 '12 at 13:59
  • $\begingroup$ Yes. It's a proper etale Lie groupoid whose objects have no nontrivial automorphisms. In other words for for each object $x$ of this groupoid the group $Hom (x, x)$ of arrows from $x$ to itself is trivial. $\endgroup$ – Eugene Lerman Nov 15 '12 at 1:47
  • $\begingroup$ This looks a very great answer. But there are many advanced words and I don't thoroughly understand. In particular, I feel confused about the following: usually, people construct a manifold from its open cover $\{U_i\}$ requiring cocycle conditions, i.e. $\phi_{ij}\circ \phi_{jk}=\phi_{ik}$ where $\phi_{ij}$ is the transition map. So, this should be hidden somewhere in your abstract answer, but I have no idea about it. $\endgroup$ – Hang Sep 7 '18 at 2:22

A counterexample in which your first three conditions hold is the following: take two copies of the real line and glue them along the open subset $\mathbb{R}^\ast$. This can be realized as the colimit of the diagram $$ \mathbb{R}^\ast \sqcup \mathbb{R}^\ast \rightrightarrows \mathbb{R} \sqcup \mathbb{R} $$ where the first arrow is the canonical injection of the first factor into the first factor and the second into the second, and the second arrow switches the two pieces. The resulting topological spaces is the famous line with doubled origin, which is evidently non-Hausdorff.

I do not know if there is a condition you could impose to bar this example, but it does not seem probable to me that one exists.

EDIT: This is in response to Colin's comment that connectedness precludes the previous example. Take $$\mathbb{C}^\ast \rightrightarrows \mathbb{C}$$ with the first arrow being the inclusion and the second one the composition of the latter with reflection about the real line. The colimit is the closed half-plane, which is a manifold with boundary. Also, by requiring connectedness you lose coproducts!

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  • $\begingroup$ I see Todd Trimble already noticed in his comment that coequalizers would be a problem. $\endgroup$ – Alberto García-Raboso Sep 13 '10 at 13:48
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    $\begingroup$ I think another problematic example would be trying to take the coequalizer of two maps $\mathbb{R} \to \mathbb{R}$, where one is the identity and the other is smooth, increasing, and equal to the identity everywhere but an open line segment. I'd have to think hard to see if type of example is really problematic, but on the face of it, it looks as though one is forced to make some uncomfortable identifications. All that being said, it could be that the category of manifolds and all smooth maps does have some significant colimits. An interesting test case is whether idempotents split. $\endgroup$ – Todd Trimble Sep 13 '10 at 14:02
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    $\begingroup$ Actually the situation I was really concerned about is the category of complex manifolds and holomorphic maps. In this case, analyticity will rule out the coequalizer example Todd gave. On the positive side, are there some references that work out those colimits that exist? $\endgroup$ – user2529 Sep 13 '10 at 14:26
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    $\begingroup$ @Colin: for the category of complex manifolds and holomorphic maps, I think the following works: $\mathbb{C}^\ast \rightrightarrows \mathbb{C}$ with the inclusion and the inclusion followed by $z \mapsto 1/z$. The image in the quotient of the points $z=1$ and $z=-1$ are problematic. $\endgroup$ – Alberto García-Raboso Sep 13 '10 at 14:38
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    $\begingroup$ Colin, yes there are open maps which are nonidentity idempotents (take the projection of the disjoint union of two lines onto one, and then follow with one of the inclusions of one line into two). There are no nontrivial idempotents that are open inclusions however. I'm inclined to return to the question in the full category of manifolds (and similar question for the category of analytic manifolds). My instinct is that idempotents split in either category, and that this would be an easy local computation. Hopefully more time for this later (unless someone else gets there first). $\endgroup$ – Todd Trimble Sep 13 '10 at 15:00

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