Problem: Every morphism of Principal G-Bundles is an isomorphism.
First I want too say that I understand how to do this problem by explicity showing the map is homeomorphism. What I am confused about is the alternative solution:
I understand that one first assumes the principal G-bundles are product bundles. AND I DO understand that locally principal G-bundles are product. But I don't understand how then one can deduce the global case from the local one as in proposition 2.1 in
A morphism of principal $G$-bundles is an isomorphism if and only if it is an isomorphism on each fibre (you can explicitly write down an inverse if this is the case). So it suffices to check that a morphism $\varphi:P\to Q$ between principal $G$-bundles is an isomorphism on each fibre, to prove that $\varphi$ is an isomorphism.
Let $x\in B$ and $U\subset B$ sufficiently small so that $P|_U\cong U\times G\cong Q|_U$. Then we get $\varphi|_U:P|_U\to Q|_U$, which is an isomorphism. Hence, $\varphi_x:\pi_P^{-1}(x)\to\pi^{-1}_Q(x)$ is an isomorphism. This implies $\varphi$ is an isomorphism.
But really, the reason why $\varphi$ is an isomorphism if and only if the induced map on the fibres is an isomorphism, is the same reason why $\varphi$ is an isomorphism if and only if each $\varphi|_U$ for $U\subset B$ and $P|_U\cong U\times G$ is an isomorphism. So if you understand why $\varphi$ is an isomorphism if and only if the map on all the fibres is an isomorphism (which is shown in the post I linked to in the comments), then it is really not clear to me why what I said above would help you. Conversely, I'm not sure what else it is that you're asking, if the post that I linked to does not answer your question.