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.

I'd like to recast Reid's (excellent) answer slightly. The essence of it is the following principle:
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:
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. 


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 $M$. 

