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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.

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1 Answer 1

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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.

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