In the category of smooth real manifolds, do all small colimits exist? In other words, is this category smallcocomplete? I can see that computing pushouts in the category of topological spaces of smooth manifolds need not be manifolds, but this is not a proof.

5$\begingroup$ @John  For this reason second countable is not a good requirement. It is better to require paracompact, which is alos a local property, unlike second countable. Each component of a paracompact manifold will be second countable, but there may be arbitrarily many components. $\endgroup$ – Chris SchommerPries Mar 23 '10 at 16:43

2$\begingroup$ I'm requiring paracompact but not second couontable in my definition, in case anybody was wondering. $\endgroup$ – Glen M Wilson Mar 23 '10 at 16:53

2$\begingroup$ Also, note that some of the colimits that do exist in Man are different from the corresponding colimits taken in Top. $\endgroup$ – Mariano SuárezÁlvarez Mar 23 '10 at 17:47

3$\begingroup$ I have to take back my assertion that paracompact is a local condition. It is not quite a local condition, it turns out. Sorry for that. It is still preferable to second countable though, as everything you want to use second countable for really only depends on paracompact. $\endgroup$ – Chris SchommerPries Mar 24 '10 at 18:50

4$\begingroup$ (I like to think of paracompactness as being a bridge between local and global. It says that we can find a neighbourhood of each point intersecting only a finite number in our cover, that's a local condition, but it also says that we can do this simultaneously over the whole space, that's a global condition.) $\endgroup$ – Loop Space Mar 27 '10 at 2:58
I'd like to recast Reid's (excellent) answer slightly. The essence of it is the following principle:
To show that a limit or colimit doesn't exist in some category, embed your category in one where limits or colimits do exist and find some diagram in the original category whose colimit in the larger category does not lie in the image of the embedding.
The point is, it's usually much easier to show that an object $X$ of $\mathcal{D}$ is not an object of $\mathcal{C}$ than it is to show that $\mathcal{C}$ has nothing that looks like $X$. For a simpler analogy, think of the difference between proving that $(0,1)$ is not complete versus proving that $(0,1) \subseteq \mathbb{R}$ is not closed. The essence is the same, but the latter always seems to me to be a lot easier to grasp.
Back to the principle. As stated, it's not quite strong enough. You need a condition on the embedding:
Make sure that your embedding preserves those limits or colimits that already exist.
Again, by analogy: to prove that a metric space $X$ is not complete, we need a continuous map from $X$ to a complete space with nonclosed image. An arbitrary map won't do.
Back to the case in hand. As the functor $M \mapsto C^\infty(M,\mathbb{R})$ is a (contravariantly) representable embedding, it preserves colimits and so is suitable for the argument to go through.
However, it does not preserve limits so if you asked the corresponding question about limits, you'd need a different embedding. It turns out, though, that there is a complete and cocomplete category in which the category of manifolds embeds preserving all limits and colimits. That is the category of Hausdorff Froelicher spaces. Froelicher spaces may feel a little more topological than algebras so for those who, like myself, prefer topology to algebra, here's a recasting of Reid's answer using (Hausdorff) Froelicher spaces.
The key thing is that a Froelicher space is completely determined by either the smooth functions from it to $\mathbb{R}$ or the smooth curves in it (i.e. smooth functions from $\mathbb{R}$).
We take the same colimit: the pushout of
$$ \begin{matrix} \{0\} &\to& \mathbb{R}\\ \downarrow \\ \mathbb{R} \end{matrix} $$
We shall show that it is the union of the $x$ and $y$ axes in $\mathbb{R}^2$, which is clearly not a manifold.
Let us write the colimit as $X$. First, we define a smooth function $F \colon X \to \mathbb{R}^2$. It is the obvious one: it sends the first copy of $\mathbb{R}$ to the $x$axis and the second copy to the $y$axis. As these two functions agree on $\{0\}$, this is a welldefined smooth function.
We want to show that this is an initial map. One sufficient (but not necessary) condition for this is that every smooth function $f \colon X \to \mathbb{R}$ factors through $F$.
As Reid says, a smooth function $f \colon X \to \mathbb{R}$ consists of two smooth functions $f_1, f_2 \colon \mathbb{R} \to \mathbb{R}$ satisfying $f_1(0) = f_2(0)$. Let $g \colon \mathbb{R}^2 \to \mathbb{R}$ be the function $g(x,y) = f_1(x) + f_2(y)  f_1(0)$. This is smooth and we have $g(x,0) = f_1(x) + f_2(0)  f_1(0) = f_1(x)$ and, similarly, $g(0,y) = f_2(y)$. Thus $g \circ F = f$ and so every function $X \to \mathbb{R}$ factors through the inclusion $X \to \mathbb{R}^2$. Hence the inclusion $X \to \mathbb{R}^2$ is initial. Thus we can identify $X$ with its image, that being the union of the two axes.
As I said, this is merely a recasting of Reid's answer. I post it partly to make it more topological in feel, but mainly to expose the general principle which Reid uses.

$\begingroup$ Andrew, this is a great explanation! Thank you for elaborating on the general method behind showing this; it reinforced my understanding of Reid's solution (which is very intriguing). Your method using Frölicher spaces is very nice. P.S.: I'm still slightly confused as to what goes into computing $dim_{\mathbb{R}} I^{n+1}/I^{n}$, but I think it might just take some elbow grease to get it to work. I hate to belabor the point, but if this computation isn't all that trivial, I'd appreciate a literature reference. If not, just ignore me! =) $\endgroup$ – Glen M Wilson Mar 27 '10 at 1:28

$\begingroup$ One of my reasons for recasting Reid's answer was that that step seemed very opaque to me as I don't think well "algebraically" about manifolds! Once I'd worked out what was happening for myself, I figured it worth posting in case anyone else was like me. So I can't give you a reference for that bit  you should ask Reid and Chris for that! $\endgroup$ – Loop Space Mar 27 '10 at 3:02

$\begingroup$ By "As the functor
$M\mapsto C^\infty(M)$
[...] preserves colimits and so is suitable for the argument to go through." you mean that it takes them to limits, right? Anyhow, +1 and thanks for making me wonder about Fröhlicher spaces. $\endgroup$ – user2146 Mar 27 '10 at 9:39 
1$\begingroup$ Whoops! Yes. I suppose I could rescue it by considering it as a function in to C^inftyrings^OP. There's probably some technical term for this property as well (a contravariant functor taking colimits to limits or viceversa). $\endgroup$ – Loop Space Mar 27 '10 at 12:37

3$\begingroup$ The calculation of the dimension of I^n/I^{n+1} is a standard part of the theory of jets and jet bundles. It is the dimension of the vector space of those functions which vanish up to order (n1) modulo those which vanish up to order n. I first learned about this from the book "Stable Mappings and Their Singularities" by Golubitsky and Guillemin. I suspect that there are many other references out there, too. $\endgroup$ – Chris SchommerPries Mar 29 '10 at 12:47
I'll show that the pushout that glues two copies of $\mathbb{R}$ at the origin does not exist in Man. Suppose for the sake of contradiction that it did; call the resulting manifold $M$, and the common image of the origins $x \in M$. The real line $\mathbb{R}$ is a ring object in Man, and it represents the functor $X \mapsto C^\infty(X)$. So, we learn that as a ring, $C^\infty(M)$ consists of pairs of functions on the real line with the same value at the origin. Similarly we can identify the ideal $I$ in $C^\infty(M)$ of functions that vanish at $x$ with pairs of functions on the real line that both vanish at the origin. Now we may compute that $\dim_{\mathbb{R}} I^n/I^{n+1} = 2$ for all $n \ge 1$, which cannot happen for a point $x$ of a smooth manifold $M$.

2$\begingroup$ Good solution. I was just about to post a more or less identical solution. The gist of this proof and several variations is that you can compute the ring of functions on a colimit by it universal property. You the combine this with the fact that the ring of functions determines the (smooth) manifold (at least in finite dimensions, but that is enough for a counter example). So if your ring of functions behaves poorly then you don't have a manifold. $\endgroup$ – Chris SchommerPries Mar 23 '10 at 16:56

$\begingroup$ I understand your description of $C^{\infty}(M)$, but I am not following why $\mathrm{dim}_{\mathbb{R}} I^n/I^{n+1}=2$ gives us a contradiction. Can someone point me to a source which explains this? This looks good though! Thanks for the helpful discussion. $\endgroup$ – Glen M Wilson Mar 23 '10 at 17:17

4$\begingroup$ When n=1, this condition implies that the tangent space at x is 2dimensional. So if it were a manifold it would have to locally look like $\mathbb{R}^2$. However, an easy calculation shows that for the plane we have $dim_\mathbb{R} I^2/I^3 = 3$, generated by the 2jets of the functions $x^2$, $xy$ and $y^2$. Choosing coordinates appropriately, the actual colimit has functions whose 2jets are $x^2$ and $y^2$, but none with $xy$. $\endgroup$ – Chris SchommerPries Mar 23 '10 at 17:32

$\begingroup$ Why did you consider Iadic filtration? Was that just an arbitrary approach to obtain a contradiction or was that somehow related to my imagination of $X$ that it should look like the tangent cone of a cusp at its singularity? $\endgroup$ – user2146 Mar 27 '10 at 11:16

$\begingroup$ Hi Reid and Chris, I've finally worked out exactly what you mean with this proof. It is very nice! Thanks very much. Best, Glen $\endgroup$ – Glen M Wilson Apr 15 '10 at 2:14