Sorry for the bad title. I could not think of a better one. I've been reading about vector bundles (all vector bundles are assumed to be smooth and all vector spaces are assumed to be real) and how to construct new ones from old ones. There is a very general way which uses category theory. This is how it goes:
Define $\mathscr{V}$ to be the category whose objects are finite-dimensional (real) vector spaces and whose morphisms are linear isomorphisms. Let $\mathcal{F}:\mathscr{V}\times\mathscr{V}\rightarrow\mathscr{V}$ be a functor. Given two $n$ -dimensional vector spaces $V$ and $V'$, denote the space of all isomorphisms from $V$ to $V'$ by $\mbox{Iso}\left(V,V'\right)$. By fixing bases for $V$ and $V'$, we can identify $\mbox{Iso}\left(V,V'\right)$ with $\mbox{GL}\left(n,\mathbb{R}\right)$. Thus $\mbox{Iso}\left(V,V'\right)$ is an $n^{2}$-manifold. We say that $\mathcal{F}$ is smooth if, for all finite-dimensional vector spaces $V$ , $V'$, $W$, and $W'$ with $\mbox{dim}V=\mbox{dim}V'$ and $\mbox{dim}W=\mbox{dim}W'$, the map $$\mathcal{F}:\mbox{Iso}\left(V,V'\right)\times\mbox{Iso}\left(W,W'\right)\rightarrow\mbox{Iso}\left(\mathcal{F}\left(V,W\right),\mathcal{F}\left(V',W'\right)\right)$$ defined by $\left(f,g\right)\longmapsto\mathcal{F}\left(f,g\right)$ is smooth.
The books I am using claim the following:
Given two vector bundles $\pi:E\rightarrow M$ and $\xi:F\rightarrow M$ with the same base space and a smooth functor $\mathcal{F}:\mathscr{V}\times\mathscr{V}\rightarrow\mathscr{V}$, there exists a vector bundle $\mathcal{F}\left(E,F\right)$ with base space $M$ whose fiber at $p\in M$ is $\mathcal{F}\left(E_{p},F_{p}\right)$.
I am only consider vector bundles with constant rank (all of its fibers have the same fixed dimension). So for $\mathcal{F}\left(E,F\right)$ to have constant rank, the following must hold:
Let $\mbox{Vec}\left(n\right)$ be the set of all $n$-dimensional vector spaces and let $\mbox{Vec}\left(m\right)$ be the set of all $m$-dimensional vector spaces. Then the map defined by $\left(V,W\right)\longmapsto\mbox{dim}\mathcal{F}\left(V,W\right)$ for all $V\in\mbox{Vec}\left(n\right)$, $W\in\mbox{Vec}\left(m\right)$ is constant.
I am unsure if the result is true for any such smooth functor. I've read several textbooks and none of them mention this condition. This leads me to believe that it is true, but I have no idea how to prove it.
This is true for any functor, smoothness is not needed. The point is that two finite-dimensional vector spaces over a field are isomorphic if and only if they have the same dimension. On the other hand, the definition of a functor is set up in such a way that is preserves isomorphism. Thus for any functor $\mathcal F$ mapping finite dimensional vector spaces to finite dimensional vector spaces $\dim(\mathcal F(V))$ depends only on $\dim(V)$. This similarly applies to bifunctors, which is what you actually need.