Disclaimer: It's heavily related to my old question : Visualizing the Topology of a Vector Bundle but I wanted to open a new question because the former had already got an answer and this time my doubt is slightly different

Every definition I've encountered of Vector bundle $(E,\pi, B)$ ($E$ is the total space, $B$ the base one, and $\pi \colon E \to B$ the projection) required that $E$ is a topological space. But, as stated in the linked question, in fact we care only about the topology induced by the trivializations and the local isomorphisms.

So the question is why don't we define a vector bundle as a space whose topology is generated by the image of open sets (contained in $B\times \mathbb{F}^n$) through the local isomorphisms?

Why mathematicians defined $E$ as "only" a Topological space (with a topology that may be finer than the one constructed above)? Is there the risk of the existence of an open that is not the union of open induced as above (and maybe this open can cause trouble in some proofs)?


1 Answer 1


Your definition would work: it's a matter of taste. In your version we would say a vector bundle is a set $E$ equipped with a function $p:E\to X$ for $X$ a topological space, a vector space structure $V_x$ on $p^{-1}(x)$ for each $x\in X$, and a topology $\tau$ generated by the condition that a certain collection of maps $\phi_i:U_i\to U_i\cap X\times \mathbb{R}^n$ be fiberwise-linear homeomorphisms, where $E=\cup U_i$ and $U_i\cap X$ is open.

In the usual definition we bake the topology in as given: but that the $\phi_i$ should be homeomorphisms implies any topology $\tau$ on $E$ making $(E,\tau',p,\phi_i,V_x)$ a vector bundle must be the one induced as above.

This is because two topological spaces are homeomorphic if and only if there exists a bijection $f$ between them which is a local homeomorphism, i.e. a homeomorphism on some neighborhoods $U\ni x\to f(x)\in V$ for every $x$ in the domain. But the identity map $I:(E,\tau')\to (E,\tau)$ is a bijection which is a homeomorphism on the $U_i$ since $\phi_i I\phi_i^{-1}: (U_i\cap X)\times \mathbb{R}^n\to(U_i\cap X)\times \mathbb{R}^n$ is a homeomorphism, in fact, the identity map. Then $I=\phi_i^{-1}(\phi_i I\phi_i^{-1})\phi_i$ is a composition of homeomorphisms and thus a homeomorphism itself.

The upshot is that the two definitions are exactly equivalent. In practice we may know something about the total space of a vector bundle as a space beforehand, so that it's natural to mention the topology before describing the trivializations; but this is entirely optional.

  • 1
    $\begingroup$ such a great answer! I'm working over it a little and eventually accept it! Didn't think it was so easy. The local to global condition of an homeomorphism just doesn't want to stick in my head :(. Thanks again! $\endgroup$
    – Riccardo
    May 11, 2014 at 21:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .