To define the tensor product of vector bundles $\xi_1$ and $\xi_2$ over base $B$, Milnor-Stasheff's Characteristic Classes takes the space $\sqcup_{b \in B} F_b(\xi_1) \otimes F_b(\xi_2)$ and topologizes it (uniquely) so that we have local trivializations $\phi: U \times (\mathbb{R}^{n_1} \otimes \mathbb{R}^{n_2}) \to \pi^{-1}(U)$. I would like to prove that this construction satisfies the following universal property:
There is a bilinear morphism $h: \xi_1 \oplus \xi_2 \to \xi_1 \otimes \xi_2$ such that any other bilinear morphism $\, f : \xi_1 \oplus \xi_2 \to \eta$ factors uniquely as $f = \tilde f \circ h$, for some $\tilde f: \xi_1 \otimes \xi_2 \to \eta$.
Thanks!
My messy attempt, feel free to skip: We can take $h$ to be $(e_1,e_2) \mapsto e_1 \otimes e_2$, which is clearly bilinear and respects fibers. Given local trivializations $\phi_i: U \times \mathbb{R}^{n_i} \to \pi_i^{-1}(U)$, this map locally sends $(\phi_1^{-1}(e_1),\phi_2^{-1}(e_2))\in \{b\}\times(\mathbb{R}^{n_1} \oplus \mathbb{R}^{n_2})$ to $(\phi_1 \otimes \phi_2)^{-1}(e_1 \otimes e_2)=\phi_1^{-1}(e_1)\otimes \phi_2^{-1}(e_2)$. Since this is the natural map $\mathbb{R}^{n_1} \oplus \mathbb{R}^{n_2} \to \mathbb{R}^{n_1} \otimes\mathbb{R}^{n_2}$ at all point $b\in B$, we see that $h$ is continuous.
Bilinearity implies that $f$ maps all of $h^{-1}(e_1 \otimes e_2)$ to a single point $x \in F_b(\eta)$; we define $\tilde f(e_1 \otimes e_2)= x$. This definition on simple tensors extends to a map $\xi_1 \otimes \xi_2 \to \eta$ that is linear on fibers, and we need only verify continuity. I think it suffices to consider the local model of a bilinear morphism $f:U \times (\mathbb{R}^{n_1} \oplus \mathbb{R}^{n_2}) \to U \times \mathbb{R}^m$. Let $\{x_i\},\{y_i\},$ and $\{z_{l}\}$ be standard bases for $\mathbb{R}^{n_1}$, $\mathbb{R}^{n_2}$, and $\mathbb{R}^{m}$, respectively. At each point $b\in U$, $f$ is a bilinear map $\mathbb{R}^{n_1} \oplus \mathbb{R}^{n_2} \to \mathbb{R}^m$, i.e. a collection of functions $f^{l}_{ij}: U \to \mathbb{R}$ such that $$f\left(b,\sum_i \alpha_i x_i,\sum_j \beta_j y_j\right)=\sum_ {{l}=1}^m \left( \sum_{ij} f^{l}_{ij}(b) \alpha_i \beta_j\right)z^{l}.$$ Continuity implies that each $f^{l}_{ij}$ varies continuously with $b \in U$. Thus $f$ determines a matrix with $m$ rows and $n_1 n_2$ columns whose entries vary continuously in $U$. Using the bundle isomorphism $U \times \mathbb{R}^{n_1 n_2} \cong U \times (\mathbb{R}^{n_1} \otimes \mathbb{R}^{n_2})$ determined by our choice of bases, we obtain our map $\tilde f$. By inspection, this definition of $\tilde f$ is seen to agree with the above definition.