Isomorphic bundles, base space the same? (SW) I want to show:

If $\xi$ is isomorphic to $\eta$ then $w_i(\xi)=w_i(\eta)$

Does isomorphic here mean isomorphic bundles over the same base space? I.e. if I have an isomorphism then:
I know that I have $\psi$ is a homeomorphism, and that $\pi_1^{-1}(b)\cong \pi_2^{-1}(b)$ is an isomorphism for all $b$. Firstly, it seems then I have that $\pi_1^{-1}(b),\pi_2^{-1}(b)$ and hence that these have the same base space (i.e. $b\in B(\xi),b\in B(\eta)$).
\begin{align}
\begin{matrix}
E(\xi)&\stackrel{\psi}\longrightarrow&E(\eta)\\
\!\!\!\!\!\!\!\pi_1\downarrow&&  \quad\downarrow\pi_2\\
B(\xi)&\dashrightarrow&B(\eta)
\end{matrix}
\end{align}
Anyway, with that in mind, $\pi_2(\psi(\pi_1^{-1}(b)))=\pi_2(\pi_2^{-1}(b))=b$ tells us that $f:B(\xi)\to B(\eta)$ is the identity map, relying on $\pi_1^{-1}(b)\cong \pi_2^{-1}(b)$. Then the induced homomorphism is the identity homomorphism on the cohomology and we are done by naturality.

Yes, we normally talk about "isomorphism of bundles" only for bundles on a fixed base space.  This means that in the diagram you've written, $B(\xi)=B(\eta)$ and the diagram should commute with the bottom map being the identity map.
