Counterexample: A bundle map which is not a pullback

Question

I'm trying to find a morphism of vector bundles (aka. bundle map) that is not a pullback, preferably with compact base spaces. I'm pretty sure, I saw such example once, but I can't reconstruct it.

Answer 1

Edit: $\require{AMScd}$

No such counterexample exists. We will take as our definition of bundle map the following.

A bundle map $f$ between bundles $p \colon E \to B$ and $p' \colon E' \to B'$ is a continuous function $f \colon E \to E'$ which restricts to an isomorphism on each fiber. This then induces a map on the base spaces $\bar{f} \colon B \to B'$ making the diagram \begin{CD} E@>{f}>> E';\\ @VpVV @Vp'VV \\ B@>{\bar{f}}>> B' \end{CD} commute.

Now we claim that $p \colon E \to B$ is isomorphic to $\bar{f}^* E' \to B$. Recall that the total space $\bar{f}^*E'$ takes the form $\bar{f}^*E' = \{(b,e'\} \in B \times E' \mid \bar{f}(b)=p'(e')\}$. Consider the map \begin{align*} g \colon E &\to \bar{f}^* E'\\ e &\mapsto (p(e),f(e)). \end{align*} We claim this map $g$ is a continuous function which maps each fiber isomorphically onto the corresponding fiber (over the same base space $B$), hence must be a homeomorphism, and so $E$ was in fact the pullback of $\bar{f}$ to begin with.

Failed Counterexample Before Edit:

Let's take the trivial vector bundle over a point $\mathbb{R} \to *$ and a non-trivial bundle, say the mobius band bundle $M \to S^1$. We of course have a map of spaces $S^1 \to *$, but is this map covered by a bundle map? I believe it is. So once we have a bundle map from the mobius band bundle to the trivial bundle it can't be a pullback since then this would imply the mobius band bundle is trivial.