Consider a countable family of finite-rank vector bundles $V\_k$ over a finite-dimensional smooth manifold $M$. The direct limit of such a family is still a topological vector bundle even though it may have infinite rank. 


The first question is: Is there a natural structure of smooth manifold on the total space of $\varinjlim V\_k$? 


And the second one: Is it true, that the functor of smooth sections commutes with direct limit? I.e, is it true that
$$\mathcal{C}^\infty(M,\varinjlim V\_k) = \varinjlim \mathcal{C}^\infty(M,V\_k)$$


Or turning things upside down -- is there a choice of smooth structure such that the above equation holds? If so, how does it look like?



Does direct limit commute with functor of smooth sections?
The answers are: Yes and No-but-yes-if-M-is-compact.


1. Kriegl and Michor's *A Convenient Setting for Global Analysis* describes how to put a smooth structure on an arbitrary locally convex topological vector space, say $V$, by looking first at the smooth curves in $V$ (these can be unambiguously defined). This works for $V = \lim V\_k$ where the $V\_k$ are finite dimensional. As your family is countable, this is just $\sum\_{\mathbb{N}} \mathbb{R}$. Smooth curves are continuous, and therefore an important property of this structure is that if $c \colon \mathbb{R} \to \sum\_{\mathbb{N}} \mathbb{R}$ is smooth then $c([a,b])$ is contained in a finite dimensional subspace.
2. And that's really the key to the second answer. If $M$ is compact, then its image lies in a finite dimensional subspace of $\lim V\_k$, whence inside one of the $V\_k$. If $M$ is not compact then it will be possible to find a smooth map $M \to \lim\_k V\_k$ which does not lie in any subspace and so the limit will not commute with the mapping space construction. Note that it is enough to show that such a map exists for $M = \mathbb{R}$, whereupon you take curves $\alpha\_k \colon [0,1] \to V\_k$ which map to $0$ on their endpoints, are infinitely slow there, and such that the image of $\alpha\_k$ contains a basis for $V\_k$. Then concatenate these paths together to get a smooth path $\alpha \colon \mathbb{R} \to \lim V\_k$ with image not contained in any finite dimensional subspace.


