If you have a smooth map $F: \mathbb{R}^n \to \mathbb{R}^m$ such that $F = E_v \circ \tau$ where $\tau: \mathbb{R}^n \to L(V ;\mathbb{R}^m)$ and $E_v:L(V;\mathbb{R}^m) \to \mathbb{R}^m$ is the map taking $A \mapsto Av$ where $v$ is in the vector space $V$, can you show that $\tau$ is smooth?
I feel like it must be because $E_v$ is a linear map, so $d_x(F) = d_x( E_v \circ \tau)$. If $\tau$ were explicitly differentiable then the chain rule would yield $d_x( E_v \circ \tau) = d_{\tau(x)}(E_v)\circ d_{x}(\tau) = E_v \circ d_{x}(\tau)$. The context for this problem is showing that a bundlemap map $G: U \times V_0 \to \mathbb{R}^m \times V_0$ where $G(x,v) = (f(x),\tau(x)v)$ is smooth iff $f:U \to \mathbb{R}^m$ and $\tau$ (as above) are smooth. I realize how to do this if we can pick a basis for $V_0$, but my professor said he would prefer us to do this abstractly.