I'm in doubt about the topology of maps between fibres of vector bundles.


Consider $E$ and $F$ vector bundles and the set of all linear maps from a fibre of $E$ to a fibre of $F$, ie, the set of all linear maps $T:E\_x \rightarrow F\_y$, where $E\_x$ is the fiber over $x$ and $F\_y$ is the fiber over $y$.


I want to know how to define the topology of this set.


I need this topology for this question: Consider $f: E \rightarrow F$, a map that preserves each fiber and its restriction to each fibre, $f\_x : E\_x \rightarrow F\_y$, is differentiable. The differential of $f\_x$ calculated in a vector $v \in E\_x$ is the linear maps $df\_x(v):E\_x \rightarrow F\_y$. I want to say that $f$ is a $C^1$ map if the function $v \in E \rightarrow df\_{\pi(v)} (v)$ is continuous. And for this I need a topology for the set defined above.


Does anybody know how to define the topology? What does it mean that two of these maps are close to each other?


Note that $df\_x (v)$ is not necessarily a homomorphism, because this is only defined over the fiber that contains $v$.



Topology of maps between fibers of vector bundles
If both bundles were trivial, say $X\times \mathbb{R}^n\rightarrow X$ and $Y\times \mathbb{R}^n\rightarrow Y$ one could just take $X\times Y\times M(m,n,\mathbb{R})$.
Different choices of trivializations should give homeomorphic spaces, so this topology seems to be right.


In general the bundles are just locally trivial. However this is enough to write down a basis for the topology. 


I don't see, what nice properties this topology might have but it seems to be the canonical choice.


EDIT: I think one can reduce this to classical vector bundle constructions in the following way. Given two vector bundles $p\_i:E\_i\rightarrow X\_i$ (for $i=1,2$). Let $pr\_i:X\_1\times X\_2\rightarrow X\_i$ denote the projections. Then the desired bundle is just Hom$(pr\_1^\*(p\_2,E\_2,X\_2),pr\_2^\*(p\_1,E\_1,X\_1))$.


