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  1. 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

      enter image description here

      commutes. Moreover, the pullback $(P, p_1, p_2)$ must be universal with respect to this diagram.

  2. Also, is it possible to define pushforward/pushout in terms of composition of functions?

Thanks and regards!

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Have a look at this question of mine:… – BBischof Apr 10 '11 at 14:36
Consider $f = g \circ \alpha$. The universal property implies that the corresponding pullback satisfies (using Wikipedia notation current at the time of writing) $P = \text{dom } \alpha$, $p_1 = id$, $p_2 = \alpha$, so that $\alpha$ is the pullback of $f = g \circ \alpha$ along $\alpha$. – S Huntsman Apr 29 '14 at 14:26
up vote 4 down vote accepted

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.

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Thanks! Could you explain the concepts of pullback and pushforward in terms of composition of functions only? – Tim Apr 10 '11 at 21:10

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

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Thanks! Can the relation be explained without referring to differential geometry, just based on the two concepts themselves? – Tim Apr 10 '11 at 7:33
I am not that well-versed in category theory, but it looks like saying that the diagram commutes means that the composition of f and p1 always gives the same result as the composition of g and p2; if p1 is invertible, then this means g is the pullback of f by the composition of p2 and the inverse of p1, in something like the earlier sense (if all of the morphisms are functions, then it is exactly like the earlier sense). – James Edward Lewis II Apr 10 '11 at 7:55
Thanks! Also is it possible to define pushforward/pushout in terms of composition of functions? – Tim Apr 10 '11 at 8:02
If g is the pullback of f by y, this means $g=f\circ y$; if y is invertible, then f is the pushforward of g by y, $f=g\circ y^{-1}$; I based this notion on a similar relationship between pushforward and pullback in differential geometry:… – James Edward Lewis II Apr 10 '11 at 9:28

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