I often seen a subscript after a Cartesian product, e.g. $A \times_\kappa B$, for example in this Wikipedia article about the spin bundle. What is the meaning of this generalised Cartesian product? Thanks.

  • $\begingroup$ It is know as a pullback, or as a fiber(ed) product. Searching those terms will help you. $\endgroup$ – Alex G. Nov 2 '15 at 12:27
  • $\begingroup$ Thanks. I just had no idea how to search for it. I will look that up. $\endgroup$ – Alex Nov 2 '15 at 12:29

This usually denotes a fibered product or pullback, a categorial generalisation of the ordinary product.

As you tagged (general-topology), I will describe the pullback in the category of topological spaces, but let's begin with the categorical description: Suppose we are given three spaces $X$, $Y$ and $Z$ and two continuous maps, $f \colon X \to Z$ and $g \colon Y \to Z$, think of them as follows: $$ \begin{matrix} && X\\ & &\downarrow f\\ Y & \stackrel g\to & Z \end{matrix} $$ Then, the pullback $X \times_Z Y$ (along $f$ and $g$) is given by a topological space $X \times_Z Y$ and to continuous maps (called projections), $\pi_X \colon \def\p{X \times_Z Y}\p \to X$ and $\pi_Y \colon \p\to Y$, such that $$ \begin{matrix} \p &\stackrel{\pi_X}\to & X\\ \downarrow \pi_Y& &\downarrow f\\ Y & \stackrel g\to & Z \end{matrix} $$ commutes and $\p$ is universal with this property, that is if we are given a space $P$ and maps $p_X \colon P \to X$ and $p_Y \colon P \to Y$, such that $$ \begin{matrix} P &\stackrel{p_X}\to & X\\ \downarrow p_Y& &\downarrow f\\ Y & \stackrel g\to & Z \end{matrix} \tag 1 $$ commutes, there is an unique continuous $i \colon P \to \p$ such that $\pi_X i = p_X$ and $\pi_Y i = p_Y$.

For topological spaces, the pullback is given by the subspace (with the subspace topology) of the ordinary product $$ X \times_Z Y := \{(x,y) \in X \times Y: f(x) = g(y) \} $$ and the projections are the usually coordinate projections (to be more exact, their restrictions). For suppose, we are given some $P$ as in (1), we may define $i \colon P \to \p$ by $i(q) = \bigl(p_X(q), p_Y(q)\bigr)$, as $p_X$ and $p_Y$ is continuous, $i$ is, as it is well-defined, since (1) commutes. $i$ is unique, since we must have $\pi_X i = p_X$ and dito for $Y$.

  • $\begingroup$ Thanks. This is a really helpful explanation. I have two further questions, though. In the article I linked, the subscript on the product is a map. What does this mean? Also, if I am given, say, $X\times_Z Y$ with no mention of the maps $f,g$ how can this be understood? $\endgroup$ – Alex Nov 2 '15 at 14:22
  • $\begingroup$ If you are given $X \times_Z Y$ without mentioning the maps, the maps have to be and are usually "obvious" from context. $\endgroup$ – martini Nov 2 '15 at 14:24

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