Let $\mathbb F$ be a field, $B$ a topological space, and consider $\operatorname{Vect}^1_{\mathbb F}(B)$, the collection of (isomorphism classes) of line bundles over $B$. There are clearly examples of non-trivial line bundles over spaces, the most obvious being the Moebius bundle over $S^1$.

In showing that $(\operatorname{Vect}^1_{\mathbb F}(B), \otimes)$ is an abelian group, I came across a slight conundrum, in that it seems like I am able to prove that every line bundle is trivial, so I am clearly making an egregious error! The proof essentially goes as such:

Let $\xi=\{p:E \to B\}$ be an $\mathbb F$-line bundle over $B$, and $\epsilon^1 = \{\hat p: B \times \mathbb F \to B\}$ the trivial line bundle. We know that for each $b \in B$, the fibre $E_b$ is isomorphic to $\mathbb F$. Since this isomorphism is not canonical, let $q_b: E_b \to \mathbb F$ be a choice of isomorphism for each point $b \in B$ and define $\phi: \xi \to \epsilon^1$ as the map which takes points $(b,v) \mapsto (b,q_b(v))$.

Now by standard results in vector-bundle theory, a vector-bundle morphism $ \eta_1 \to \eta_2$ (over the same base) is a vector-bundle isomorphism if and only if the corresponding map on the total spaces $E(\eta_1) \to E(\eta_2)$ restricts to a linear isomorphism on each fibre. The map $\phi$ defined above is a $B$-bundle map as it respects the projections; that is, $\hat p \phi = p$. It also seems to be an isomorphism on each fibre, so by theorem should be a bundle isomorphism.

The proof is obviously false. In particular, it seems to have somehow forgotten any twist which might occur on $\xi$, which really suggests to me that the problem occurs in the `gluing' of the fibres. I figure I must have missed some sort of regularity condition, but double checking Husemoller leads to no evidence as such. Where is the fallacy in the logic?

  • 4
    $\begingroup$ Note that $B$-bundle maps have to be continuous maps between the total spaces, choosing something for every fiber is not continuous. $\endgroup$
    – martini
    Commented Oct 9, 2013 at 20:36
  • $\begingroup$ This would certainly be that missing regularity condition. Intuitively it seemed that this should be the case, but my reference texts never alluded to the fact that the morphism should need to be continuous. I guess this is just something that should be obvious eh? $\endgroup$ Commented Oct 9, 2013 at 20:38
  • $\begingroup$ To those who are stumbling on this problem, something that helped a while ago is to talk about when fibre bundles are isomorphism in each category. As sets, there is an obvious bijection given above, so all bundles over a fixed base with fixed fibres are isomorphic as sets. In the category of topological spaces, one only needs to find a continuous bijection with continuous inverse. In the category of bundles, we necessarily need a bundle homomorphism with continuous bundle homomorphism inverse. $\endgroup$
    – Chris
    Commented May 26, 2023 at 19:54

1 Answer 1


A line bundle is something that looks like $U\times\mathbb F$ for sufficiently small open $U$. A morphism of fiber bundles over $B$ is defined as something that looks nicely (e.g continuous) on those $U\times \mathbb F$ patches. Review your definitions. It seems that you fell for identifying $U\times\mathbb F$ with $\{x\}\times \mathbb F$, which is canonically isomorphic, but that is indeed how the fibres are "glued" together.

  • $\begingroup$ I think my real issue was that when I reviewed my definitions, there did not seem to be any regularity conditions imposed upon the bundle map (somebody double check Husemoller and see if you can find the word continuous!). I guess this was silly of me: the bundle map should always preserve the structure of the fibres and the topological regularity of the base. Anyway, I thought the whole point was that while each fibre is isomorphic to $\mathbb F$, they are not canonically so unless the bundle is trivial. Also, wouldn't $U\times \mathbb F$ only be homotopy equivalent to $\{x\}\times\mathbb F$? $\endgroup$ Commented Oct 10, 2013 at 15:37

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