Subbundle of a pullback is not the pullback of a subbundle

Question

Suppose $p\colon E\to X$ is a vector bundle of finite rank and let $f\colon Y\to X$ be a continuous map. My professor claimed that subbundles of the pullback $f^*(E)$ are not themselves necessarily pullbacks of subbundles of $E$ along $f$. Why is this case? Can someone provide an example of when the words "subbundle" and "pullback" fail to commute in this manner?

The total space of the pullback is defined to be $$\{(y,v)\in Y\times E\colon f(y)=p(v)\}.$$ The total space of a subbundle $S$ lives inside this set. Is there no canonical way to identify bases of fibers in the pullback $f^*(E)$ with bases of fibers in $E$ itself so that the subbundle $S$ can be identified with a subbundle of $E$?

For context: I was studying Chern classes in the setting of algebraic geometry. I saw that for a vector bundle $E$ on a scheme $X$, a certain pullback of the vector bundle, $f^*(E)$, admits a "Chern decomposition" into a product involving the Chern roots. I believed that this will also imply there exists a Chern decomposition of $E$ itself but apparently this is not the case. The obstruction to this is that subbundles of $f^*(E)$ may fail to be pullbacks of subbundles of $E$. Comments on this topic are also welcome.

Answer 1

Consider a subbundle $q\colon F\rightarrow X$ of $p\colon E\rightarrow X$, which can be decomposed as $q\colon F\hookrightarrow E\xrightarrow{p}X$. When looking at the pullback bundles $f^*E$ and $f^*F$ over $Y$, you get an induced map $f^*F\rightarrow f^*E$ due to the universal property of the pullback.

Since the whole and lower square are both pullbacks, the upper square is a pullback. Therefore $f^*F\hookrightarrow f^*E$ is injective as monomorphisms are stable under pullback. This shows, that pullbacks of subbundles are again subbundles. However, not every subbundle of the pullback bundle arises that way.

Consider the map $S^1\rightarrow S^1,z\mapsto z^n$. The pullback of the trivial bundle $\underline{\mathbb{R}^2}$ with zero twists is again the trivial bundle $\underline{\mathbb{R}^2}$ and the pullback of the Möbius bundle with one twist will be the line bundle with $n\operatorname{mod}2$ twists, so the trivial bundle for $n$ even and again the Möbius bundle for $n$ odd. So for $n$ even, the Möbius bundle is a subbundle of the pullback bundle $\underline{\mathbb{R}^2}$, but cannot arise as a pullback bundle of a subbundle of $\underline{\mathbb{R}^2}$.

Answer 2

Here is a somewhat stupid example: Take $X = \{*\}$ the one-point set, then any vector bundle $E$ on $X$ is trivial, and so is the pull-back $f^*E = \mathbb R^n \times Y$. But in general, $\mathbb R^n \times Y$ might have a non-trivial subbundle¹ $F \subset \mathbb R^m \times Y$, so $F$ cannot be a pull-back from $X$.

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¹ as in the other answer take $Y = S^1$ with the Möbius bundle $F \subset \mathbb R^2 \times S^1$.