# Morphisms between trivial vector bundles

I'm reading about $$\mathbb{K}$$-vector bundles (where $$\mathbb{K}= \mathbb{R},\mathbb{C}$$), and in one example is said that

A morphism $$T:V \times B \rightarrow W \times B$$ between trivial vector bundles over a smooth manifold $$B$$ is the same as a smooth map $$\bar{T}: B \rightarrow \text{Hom}(V,W)$$.

I've tried to prove it. Given a morphism $$T$$, one can construct a function $$\bar{T}$$ given by $$b \mapsto T|_{V \times b}$$, which is well defined since $$T$$ preserve the fibres and is linear on them. But I couldn't prove that this $$\bar{T}$$ is smooth. Here is my attempt so far:

The smooth structure on $$V$$ (dimension $$r$$) and $$W$$ (dimension $$s$$) are given by linear isomorphism $$\cong_{V}$$ and $$\cong_{W}$$, respectively. These isomorphisms induce a linear isomorphism $$\text{Hom}(V,W) \cong \text{Hom}(\mathbb{K}^r,\mathbb{K}^s)$$ given by $$L \mapsto \cong_{W} \circ L \circ \cong_{V}^{-1}$$, and one knows that $$\text{Hom}(\mathbb{K}^r,\mathbb{K}^s) \cong \mathbb{K}^{sr}$$. The composition between both isomorphisms is a smooth structure ($$\psi$$) on $$\text{Hom}(V,W)$$. With this at hand, $$\bar{T}$$ is smooth if for every $$b \in B$$, there exists a smooth chart $$(b\in U, \phi)$$ such that $$\psi \circ \bar{T} \circ \phi^{-1}$$ is smooth. However, I don't see how to connect the smoothnes of $$T$$ to fulfill this requirement.

Could you give me a hand?

Any help is appreciated.

Let $$(U, x_1, \dots, x_d)$$ be a chart on $$B$$ (with $$d = \text{dim}B$$) and on this chart the smooth map $$T$$ looks like (after fixing a basis for $$V, W$$) $$(v, x)\mapsto (A(x)v, x)$$ where $$A(x) = (A_{ij}(x))$$ is some matrix. The smoothness of $$T$$ means that each $$A_{ij}(x)$$ is a smooth function: consider $$\\{e_i\\}\times U\hookrightarrow V\times U\to W\times U\to \mathbb{K}f_j$$ where the first map is inclusion, second $$T$$ and third projection (with $$e_i$$s forming a basis for $$V$$ and $$f_j$$s one for $$W$$). This smooth map is precisely $$A_{ij}(x)$$ and the map $$U\to\text{Hom}(V,W)$$ is given by $$x\mapsto (A_{ij}(x))$$ hence is smooth.