## Pullback bundle as pullback example (category theory)

Sometimes I think I’m in an  odd variation of the game Twister except instead of colored circles, the circles are labeled with topics.  Currently I have at least one limb on TQFTs, another limb on Lie theory/representation theory, another limb on quantum mechanics, and the last limb moving back and forth across topics that come up.  Fiber bundles are an example. I know these turn up in physics (though I don’t know why yet).  I introduced them in Spheres-2.  To continue the analogy, I only moved a limb to the circle labled “Fiber Bundles” for a short while, enough to digest some information.  I’m glad I did though because they provide a nice illustration of pullbacks.

The term pullback is loaded.  I think of pullbacks of functions and differential forms in differential geometry for example.  It’s entirely possible that the category theory definition is consistent with that but I can’t see it yet.  I learned about pullbacks (and their dual: pushouts) recently watching the Catsters again on Youtube.  They have two very good videos: Pullbacks and Pushouts -1 and Pullbacks and Pushouts – 2.

The initial setup is this: you have three objects A,B, and C in your category and a pair of maps, $f:A\to C$ and $g:B\to C$.  As a diagram,

From here we can talk about the pullback of g along f or the pullback of f along g.  I think for the moment I’ll talk about the pullback of the diagram which, though I haven’t heard or read, seems appropriate.  It turns out that this construction is an example of a limit of a diagram (something incredibly cool in its own right).  The pullback of this diagram is another object D and another pair of maps $s:D\to A$ and $t:D\to B$ such that the following square commutes,

$\begin{matrix} D & \overset{t}{\longrightarrow} & B \\ \downarrow_s & & \downarrow_g \\ A & \overset{f}{\longrightarrow} & C \end{matrix}$

This construction, like many I’ve been learning about, satisfies a universal property.  This means that for any other object E and morphisms $m:E\to A$ and $n:E\to B$ there exists a unique morphism (dotted arrow) such that the resulting triangles commute,

As a verb, pullback seems appropriate at least in the sense that the maps g and f are pulled back to a common domain.  But it’s not quite the same as when you have a bunch of manifolds and functions $M\overset{f}{\longrightarrow} N\overset{g}{\longrightarrow} S$ and you think of pulling back g to M.

This is where the pullback bundle comes in.  Suppose we have a fiber bundle E over a base space B with projection $\pi : E\to B$.  Suppose we have another topological space B’ and continuous map $f:B'\to B$ then we can pull back the bundle over B to become a bundle over B’, call it E’.  We construct E’ by taking a subset of $B'\times E$ composed of all those pairs $(b', e)\in B'\times E$ such that $f(b')=\pi(e)$.  We’ll need a projection map from E’ to B’ and we define it this way $\pi'(b',e)=b'$  We also have another projection onto the other coordinate, $f':E'\to E$.  The resulting diagram,

$\begin{matrix} E' & \overset{f'}{\longrightarrow} & E \\ \downarrow_{\pi'} & & \downarrow_{\pi} \\ B' & \overset{f}{\longrightarrow} & B \end{matrix}$

commutes by construction (details are from wikipedia’s page on the pullback bundle, but note the similarities between how Cheng (Catsters) constructs pullbacks).  So what about the fibers?  Roughly, if $b\in B$ then $\pi^{-1}(b)$ is homeomorphic to the fibers, F, of E.   What about B’?  Let’s take $b'\in B'$ and consider $\pi'^{-1}(b')\subset E'$.  What is this?  It’s all the ordered pairs with first coordinate b’ and second coordinate any $e\in E$ such that $\pi(e)$ is projected to the same point as b’ is sent to under f. Those $e\in E$ collectively form a fiber (or a subset homeomorphic to the fiber F). This means that $\pi^{'-1}(b')={b'}\times\pi^{-1}(f(b'))$ which is homeomorphic (pretty sure anyways) to ${b'}\times F$ which is certainly homeomorphic to F.  In this way we pulled back the space E which fibers over B to a space E’ that fibers over B’. The following  illustration might help,