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?
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.