Product in the category of smooth vector bundles

Question

Is the product in the category of smooth vector bundles just the direct product of smooth vector bundles?

More precisely, if $\pi:E \rightarrow B$ and $\pi': E' \rightarrow B'$ are two smooth vector bundles over the manifolds $B$ and $B'$, then is the vector bundle $ \pi \times \pi' : E \times E' \rightarrow B \times B'$ a product in the category of smooth vector bundles?

To me it seems the correct choice!

Answer 1

Yes. However, let's delineate the situation to avoid confusion. There is a category whose objects are vector bundles $\pi\colon E\rightarrow B$ over a fixed base $B$ and where a morphism $(\pi\colon E\rightarrow B)\rightarrow(\pi^{\prime}\colon E^{\prime}\rightarrow B)$ is a smooth map $F\colon E\rightarrow E^{\prime}$, such that $\pi^{\prime}\circ F=\pi$ and $F$ is fiber-wise linear. The product of $(\pi_1\colon E_1\rightarrow B)$ and $(\pi_2\colon E_2\rightarrow B)$ in this category is the fiber-wise direct product $\pi\colon E_1\times E_2\rightarrow B$ (which is the same as the direct sum in this case), the canonical projections given as the fiber-wise canonical projections of vector spaces. I assume this as known.

Then, there is also the category you are interested in, the category of smooth vector bundles not over a fixed base. The objects are smooth vector bundles and a morphism $(\pi\colon E\rightarrow B)\rightarrow(\pi^{\prime}\colon E^{\prime}\rightarrow B^{\prime})$ is a smooth map $f\colon B\rightarrow B^{\prime}$ and a smooth map $F\colon E\rightarrow E^{\prime}$, such that $\pi^{\prime}\circ F=f\circ \pi$ and $F$ is fiber-wise linear. A very convenient perspective to take here is that this is actually the same data as a smooth map $f\colon B\rightarrow B'$ and a bundle map over the fixed base $B$ from $\pi\colon E\rightarrow B$ to the pullback $f^{\ast}\pi^{\prime}\colon f^{\ast}E^{\prime}\rightarrow B$.

Now, given two bundles $\pi_1\colon E_1\rightarrow B_1$ and $\pi_2\colon E_2\rightarrow B_2$, consider the bundle $\pi_1\times\pi_2\colon E_1\times E_2\rightarrow B_1\times B_2$ with the canonical projections given by the canonical projections on the base spaces and the fiber-wise canonical projections of vector spaces, and allow me to notationally suppress the projections from now on. If $p\colon T\rightarrow A$ is any bundle, a bundle map $T\rightarrow E_1\times E_2$ is given by a smooth map $f\colon A\rightarrow B_1\times B_2$, which is equivalent to giving the components $f_1\colon A\rightarrow B_1$ and $f_2\colon A\rightarrow B_2$ via the canonical projections, and then a bundle map over the fixed base $A$ from $T$ to $f^{\ast}(E_1\times E_2)$. The punchline is that, canonically, $f^{\ast}(E_1\times E_2)=f_1^{\ast}E_1\times f_2^{\ast}E_2$ as bundles over $A$. Thus, by the first paragraph, giving this bundle map is equivalent to giving the bundle maps $T\rightarrow f_1^{\ast}E_1$ and $T\rightarrow f_2^{\ast}E_2$ via the canonical projections. In total, this is thus equivalent to giving the two bundle maps $T\rightarrow E_1$ and $T\rightarrow E_2$. If you carefully chase through these identifications, you will see that this identification is indeed given by the canonical projections of the bundles, thus proving the universal property of the product.

The above remains accurate if you replace "smooth manifold" by "topological space" and "smooth vector bundle" by "topological vector bundle" everywhere. It just happens to be the case that all these constructions respect smoothness.

Lastly, two tangential remarks.

1. The forgetful functor from the category of smooth vector bundles to the category of smooth manifolds mapping a bundle to its base (alternatively, from the category of topological vector bundles to the category of topological spaces) has a left-adjoint, given by sending a smooth manifold (resp. a topological space) $X$ to the trivial rank $0$ vector bundle $X\times\mathbb{R}^0\rightarrow X$. Thus, the forgetful functor preserves products, so even a priori, the base of the product of two bundles has to be the product of the bases.

2. In the category of (smooth) vector bundles not over a fixed base, the distinction between direct sum and direct product is more relevant. There is a coproduct in this category, but it is simply the disjoint union in both base and total space. Thus, the direct sum of two bundles does not have an evident categorical interpretation anymore.

Answer 2

Maybe a small remark, but your confusion is most certainly understandable: There are two different operations that are a priori different:

1. The Whitney sum of two vector bundles $E,F$ over a base $B$ is the vector bundle $E\oplus F$ with fibres $E_b\oplus F_b$ over each point $b \in B$. This is not a categorical fibre product of topological spaces, but comes from seeing $E,F$ as elements of $\mathbf{Coh}(B)$, the abelian category (where you have bi-products) of coherent sheaves on the manifold $B$. In fancy terms, this is an algebraic operation of modules over the ring of regular (or $C^\infty$-functions if you prefer that) $\mathcal{O}_B$ of $B$ (see "algebraic geometry").
2. the second thing is the fibre product of topological spaces, where given $f:B' \rightarrow B$ and $E \rightarrow B$ smooth morphisms of manifolds you get a pullback bundle $f^*E \rightarrow B'$ with fibres $f^*E_{b'}=E_{f(b')}$ for $b' \in B'$.

Now you get the Whitney sum $E\oplus F$ over $B$ in $(1)$ from the pullback $(2)$ by taking the diagonal map $\Delta:B \rightarrow B\times B $ and the bundle $E\times F \rightarrow B\times B$ over the product (product in the Top category). Then set $E\oplus F = \Delta^*(E\times F)$ which is a thing only over $B$.