# Is $C^\infty(M,\mathbb{R})$ an ind-(smooth manifold)?

Let $\mathrm{Man}$ be the category of smooth manifolds (2nd countable, Hausdorff, no boundary, not necessarily compact) and smooth maps, and let $M$ be an object thereof. Is the presheaf $S \mapsto C^\infty(S \times M,\mathbb{R})$ on $\mathrm{Man}$ isomorphic to a filtered direct limit of representable functors?

("Yes" when $\dim(M) = 0$, possible strategy in general: If $C^\infty(M,\mathbb{R})$ is made into some kind of infinite-dimensional manifold, perhaps the finite-dimensional submanifolds of it would form a filtered category.)

• There's a natural ind-manifold structure on any infinite-dimensional real vector space given by looking at its finite-dimensional subspaces. This won't represent the functor you want, though. – Qiaochu Yuan Oct 8 '15 at 17:36
• Your category of manifolds is small, so there is a necessary and sufficient condition for presheaves to be ind-representable. In this case it boils down to the following problem: given smooth maps $f_0, f_1 : X \to Y$ and a smooth function $q : Y \times M \to \mathbb{R}$ such that $q \circ f_0 = q \circ f_1$, does there exist a smooth map $g : Y \to Z$ and a smooth function $r : Z \times M \to \mathbb{R}$ such that $g \circ f_0 = g \circ f_1$ and $r \circ (g \times \mathrm{id}_M) = q$? – Zhen Lin Oct 8 '15 at 18:18
• About your strategy (last paragraph), it seems likely to me that two one-dimensional submanifolds can be chosen in such a way that any submanifold containing them both would have to be infinite-dimensional. Let one of them be a line, and let the other one cross it at a convergent infinite sequence of points, crossing in a different direction every time. – Tom Goodwillie Oct 8 '15 at 23:56