Hatcher's, tensor product of vector bundles : topology explained So this is a sumamry of the construction given in pg14 of Hatchers of the tensor of two vector bundles:

Let $p_1:E _1 \rightarrow B$, and $p_2 : E_2 \rightarrow B$. We define $E_1 \otimes E_2$ to be the disjoint union of vector spaces $p^{-1}(\chi) \otimes p_2^{-1}(\chi)$ for $\chi \in B$ with the following topology:
We take an open cover of $B$ over which both initial bundles are trivial. Choose hoemomoprhisms,
$$h_i:p_i^{-1}(U) \rightarrow U \times \mathbb{R}^{n_i}$$
for each open set $U$ in such a cover. The topology on $\tau_U$ on the set $p_1^{-1}(U) \otimes p_2^{-1}(U)$ is defined by letting the tensor product map
$$h_1 \otimes h_2: p^{-1}_1(U) \otimes p_2^{-1}(U) \rightarrow U \times (\mathbb{R}^{n_1} \otimes \mathbb{R}^{n_2})$$

I am just confused with what exactly is meant here. So then what is the topology on $E_1 \otimes E_2$? I suppose it is the colimit topology from the open cover.
 A: You are right. That is the usual procedure to topologize a "bundle" over a base space $B$ which is given by associating to each $b \in B$ a vector space $E_b$. Define $E= \bigcup_{b \in B} \{ b\} \times E_b$ = disjoint union of the fibers $E_b$ and $p : E \to B, p(b,e_b) = b$. Assume you have an open cover $\mathcal{U}$ of $B$ and for each $U \in \mathcal{U}$ an "algebraic bundle isomorphism" $h_U : p^{-1}(U) \to U \times \mathbb{R}^n$ (that is, $h_U(b,e_b) = (b, \phi_U^b(e_b))$ with linear isomorphism $\phi_U^b : E_b \to  \mathbb{R}^n$, $b \in U$). If the obvious compatibility conditions are satisfied (the transition functions between the "bundle charts" $h_U$ have to be homeomorphisms), then you can topologize $E$ by 
(1) giving each $p^{-1}(U)$ the unique topology making $h_U$ a homeomorphism
(2) giving $E$ the colimit topology from the cover $p^{-1}(U)$, $U \in \mathcal{U}$.
In your case the situation is that any two (topological) bundle charts $h_i : p_i^{-1}(U) \to U \times \mathbb{R}^{n_i}$ provide an algebraic bundle isomorphism $h_1 \otimes h_2: p^{-1}_1(U) \otimes p_2^{-1}(U) \rightarrow U \times (\mathbb{R}^{n_1} \otimes \mathbb{R}^{n_2})$. You have to check compatibility and you are done.
