The question is about the proof of the following result:
For a paracompact space $B$, the map $[B, \operatorname{Gr}_k] \to \operatorname{Vect}^k(B)$, $[f] \mapsto f^*(\gamma_k)$ is a bijection, where $\gamma_k \to \operatorname{Gr}_k$ is the universal bundle for infinite Grassmannian $\operatorname{Gr}_k$.
There is a proof in these notes. The main part is the diagram in the following picture:
$\hspace{11mm}$
Here is my question:
In the picture, there is "map $g: E \to \mathbb{R}^{\infty}$ which is linear and injective on each fiber", what does "linear and injective on each fiber" really mean? Why do we need this condition, to keep the map $f(b) = g(p^{-1}(b))$ well-defined?
As $p : E \to B$ is a vector bundle, for any $b \in B$, the fibre $E_b = p^{-1}(b)$ has the structure of a vector space. Given a map $g : E \to \mathbb{R}^{\infty}$, we can restrict it to any fibre to obtain a map $g|_{E_b} : E_b \to \mathbb{R}^{\infty}$. As $\mathbb{R}^{\infty}$ is a vector space, it makes sense to talk about the linearity of $g|_{E_b}$. If $g|_{E_b}$ is linear and injective for each $b \in B$, then we say that $g$ is linear and injective on each fibre.
As $g$ is linear on each fibre, $g(E_b)$ is a subspace of $\mathbb{R}^{\infty}$, and as $g$ is injective on each fibre, $\dim g(E_b) = \dim E_b = \operatorname{rank} E = k$. So $f(b) = g(p^{-1}(b)) = g(E_b)$ is a $k$-dimensional subspace of $\mathbb{R}^{\infty}$, and therefore $f(b) \in \operatorname{Gr}_k$.