Meaning of pullback

Solution 1:

It took me a second to figure out what the Wikipedia page was saying.

Say you have a Fiber bundle $\pi:E \rightarrow B$ and a section (which is a function $s:B\rightarrow E$ such that $\pi(s(b))=b$ for $b \in B$.)

Also, say you have a function $f:B^\prime \rightarrow B$. Then the pullback object $E^\prime$ can be defined as the set $E^\prime = \{(e,b^\prime) \in E \times B' : \pi(e)=f(b^\prime)\}$ Then you get an obvious pullback $\pi^\prime :E^\prime \rightarrow B^\prime$ defined by $\pi^\prime(e,b^\prime)=b^\prime$.

The key is that the pullback of $s$ is defined as $s^\prime:B^\prime \rightarrow E^\prime$ defined by $s^\prime(b^\prime) = (s(f(b^\prime)),b^\prime)$.

So the pullback of $s$ is (essentially) the composition of $s$ with $f$.

This pullback of $s$ can be made categorical because you have a square with top left object $B^\prime$ and bottom right $B$ defined with two paths: $B^\prime \xrightarrow{id} B^\prime \xrightarrow{f} B$ and $B^\prime \xrightarrow{s \circ f} E \xrightarrow{\pi} B$. So by the unversal property of $E^\prime$, there must be an $s^\prime:B^\prime \rightarrow E^\prime$ which, when composed with $\pi^\prime$ yields the identity.

Solution 2:

The first article you linked to gave a relationship between the two notions using concepts from differential geometry.