I was wondering if the following two meanings of pullback are related and how:
In terms of Precomposition with a function:
a function $f$ of a variable $y$, where $y$ itself is a function of another variable $x$, may be written as a function of $x$. Then $f(y(x)) \equiv g(x)$ is the pullback of $f$ by the function $y(x)$.
In the context of Category theory:
the pullback of the morphisms $f$ and $g$ consists of an object $P$ and two morphisms $p_1 : P \rightarrow X$ and $p_2 : P \rightarrow Y$ for which the diagram
commutes. Moreover, the pullback $(P, p_1, p_2)$ must be universal with respect to this diagram.
- Also, is it possible to define pushforward/pushout in terms of composition of functions?
Thanks and regards!
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.