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I'm teaching myself some differential geometry in the hope to understand gauge theory properly. In the definition of the the pullback bundle I came across a strange notation that I've never seen before. The pullback bundle was defined, with $f: M \to N$, as

\begin{equation} f^\ast \mathcal{E} \equiv (f^\ast E, f^\ast \pi, M , F), \end{equation} where the total space is given by \begin{equation} f^\ast E \equiv N \times_M E := \{ (x,z) \in N \times E | f(x) = \pi(z) \}. \end{equation}

What does the subscript on the $\times$ operator mean? Is it something to do with the operation occurring on the manifold $M$?

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    $\begingroup$ Are you familiar with Pullbacks in category theory? If not i think reading through this will help you understand the notation: en.wikipedia.org/wiki/Pullback_(category_theory) $\endgroup$ – Riquelme Jul 12 at 13:15
  • $\begingroup$ Ah perfect, thanks! $\endgroup$ – SBrents Jul 12 at 14:55
  • $\begingroup$ SBrents Can you please give a reference where you found it? $\endgroup$ – magma Jul 13 at 14:04
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This is an operation in that is sometimes referred to as "fiber product", and which has other names and a general formalization in category theory (see the comment of @Riquelme). I have never seen this particular notation for it, but it nonetheless makes a kind of sense.

The idea is that you are given two functions both with the same target $M$, in this case $f : N \to M$ and $\pi : E \to M$. You want to form a kind of restricted product of $N$ and $E$, which could be called the "product of $N$ with $E$ over $M$". This is a subset of the "true" product, and it's defined by the formula that you gave.

You can form a fiber product in many different categories: groups and topological spaces are two places where I see fiber products now and then (probably related to the fact that I am a geometric group theorist).

In your situation the fiber product is being put to work to form pullback bundles in various bundle categories, such as vector bundles or fiber bundles.

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  • $\begingroup$ Makes sense actually, thanks for the clear answer. $\endgroup$ – SBrents Jul 12 at 14:57

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