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I'm sure this is a silly question but I'm stuck at the concept of pulling back sections of a vector bundle. Let $\pi:E\to X$ be a vector bundle on a variety $X$ and $f:Y\to X$ a morphism. We have a bundle $\pi':f^\ast E\to Y$, the pullback of $E$. If we have a section $\sigma:X\to E$, which means $\pi\circ\sigma=1_X$, then I read in several places that I can pull-back this section by pre-composing it with $f$, and this gives me a section of $f^\ast E$. It seems to me that I only get an arrow $Y\to E$.

So why do we call $\sigma':=\sigma\circ f$ a section of $\pi'$?

My thoughts: call $g$ the morphism $f^\ast E\to E$. Perhaps the above means that if we start with some $y\in Y$ and we take any $z\in g^{-1}(\sigma'(y))$, then we have that $\pi'(z)=y$. Is this correct? In any case, I can only prove that $f(\pi'(z))=f(y)$, which doesn't imply $\pi'(z)=y$. However I feel like I'm really missing the point here.

Thank you.

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1 Answer 1

up vote 3 down vote accepted

The morphisms $\mathrm{Id}_Y: Y\to Y$ and the composition $\sigma\circ f: Y\to X\to E$ induce by the universal property a morphism $Y\to Y\times_X E=f^*E$. This is the pull-back of $\sigma$ to $f^*E$.

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