Whitney sum

Remarks

The map $q$ is a vector bundle of rank $n + n'$: local trivializations $p^{-1}(U) \cong U \times \R^n$ and $p'^{-1}(U) \cong U \times \R^{n'}$ combine into $q^{-1}(U) \cong U \times \R^{n + n'}$. Equivalently, $E \oplus E'$ is the pullback of the product bundle $E \times E' \to B \times B$ along the diagonal $B \to B \times B$. The Whitney sum is associative and commutative up to isomorphism, and the trivial bundle of rank $0$ is a unit.

Example: let $TS^2$ be the tangent bundle of $S^2 \subset \R^3$ and $\nu = \set{(x, t x) : x \in S^2, t \in \R}$ the normal line bundle. At each $x \in S^2$, $T_x S^2 \oplus \R x = \R^3$, and so $TS^2 \oplus \nu$ is isomorphic to the trivial bundle $S^2 \times \R^3$, although $TS^2$ itself is not trivial.

Non-example: for bundles $E \to B$ and $E' \to B'$ over different bases, the product $E \times E' \to B \times B'$ is a vector bundle of rank $n + n'$. It is not a Whitney sum, which needs a common base; it is the external direct sum.