How do I show that a fiber bundle with trivial structure group is (isomorphic to) a trivial bundle?
I say let $\phi$ and $\phi'$ be two charts over $U$. Then $\phi=\phi'$ since there is a continuous function $\theta_{\phi,\phi'}:U\to \{e\}$ such that $\phi'(u,f)=\phi(u,\theta_{\phi,\phi'}(u)\cdot f)=\phi(u,e\cdot f)=\phi(u,f)$ for all $u\in U$ and $f\in F $. How does this help me?
Let’s say we’ve covered our base space of $p: Z\rightarrow X$ with a collection of trivializations, $f_i:p^{-1} (U_i ) \rightarrow U_i \times F$ and that all the transition functions are trivial.
Recall that to get the total space back from the information of the transition functions, we glue $U_i \times F$ to $U_j \times F$ by the prescription $(x,t)=(x,\theta_{j,i}(t))$ for all pairs $(i,j)$.
Since the transition functions are trivial we are gluing via the identity. We may define the function $T: Z \rightarrow X \times F$ via $z$ goes to $z’$ where we get $z’$ by taking a trivialization $f_i$ around $p(z)$ and letting $z’=f_i(z)$.
This has an inverse defined in a similar way. The fact that this is well defined follows from the fact that we glued via the identity.