8
$\begingroup$

Usually, in a category with pullbacks, the pullback on a cospan $f : A \rightarrow C \leftarrow B : g $ is noted $ A \times_C B $. However its definition is highly dependent on the choice of arrows $f$ and $g$.

So, is there a good coherent reason to mention $C$ in the notation, instead of mentionning $(f,g)$?

Or, if not, is there a good historical reason?

$\endgroup$

2 Answers 2

4
$\begingroup$

The pullback is the categorical product in the slice category of objects over $C$.

A nice motivating case to think about is the category of schemes, where, for example, the category of schemes over a field $k$ is the slice category of objects over $\text{Spec } k$ in the category of schemes. On affine schemes the categorical product is even given by tensor product of the corresponding rings of functions over $k$ (which note that we also denote by $A \otimes_k B$ without explicitly highlighting the dependence on maps $k \to A, k \to B$).

Note that mathematical notation constantly omits dependence on extra data for the sake of brevity: for example, in representation theory it's typical to talk about a representation $V$ of a group $G$ without explicitly highlighting that all constructions involving $V$ depend on a choice of map $\rho : G \to \text{Aut}(V)$. The point is that it's understood that the symbol $V$ already implicitly refers to this data, which is usually clear from context.

$\endgroup$
7
  • $\begingroup$ So, by analogy with the representation theory example, you mean that when using pullbacks all we need is to get $any$ pullback with codomain $C$, whatever $(f,g)$ is? I think that arises from how it is mainly used when working with it. I may realize that with more experience. $\endgroup$
    – almaus
    Feb 24, 2014 at 8:04
  • 1
    $\begingroup$ @almaus: I don't understand your question, but hopefully this is an answer: in practice $f$ and $g$ will be clear from context. $\endgroup$ Feb 24, 2014 at 8:11
  • $\begingroup$ I mean, when we fetch a representation of a group, we are happy to use it as a tool to understand $G$ from within $V$, no matter what is the map $\rho$. So there should be something similar with the pullback, making it useful even if we "forgot" the exact $(f,g)$, right? $\endgroup$
    – almaus
    Feb 24, 2014 at 8:20
  • 1
    $\begingroup$ @almaus: I still don't understand your question. $f$ and $g$ haven't been forgotten; they've been omitted for brevity. $\endgroup$ Feb 24, 2014 at 18:26
  • $\begingroup$ Yes, but my quest is to grasp the understanding of why this brievity is acceptable in this case. $\endgroup$
    – almaus
    Feb 24, 2014 at 22:13
4
$\begingroup$

You are technically correct. The definition of the pullback completely depends on the arrows $f$ and $g$. One of the reasons for the prevalence of the notation $A\times_C B$ instead of something like $A\times_{f,g} B$ is that it is more aesthetically pleasing and very often does not result in ambiguity. Unless you are in the presence of another arrow with codomain $C$, the notation $A\times_C B$ can only refer to one pullback.

$\endgroup$
5
  • $\begingroup$ Ok, so is it because during mathematician job we most often work with only one pullback (and only one couple $(f,g)$) on a given codomain? I ask because I'm still learning category theory, not yet used it a lot outside the handbook. $\endgroup$
    – almaus
    Feb 24, 2014 at 7:59
  • 1
    $\begingroup$ @almaus: You are right. One only writes $A \times_{f,g} B$ instead of $A \times_C B$ if $f$ and $g$ are not understood from the context. But I also like Qiaochu's point ... $A$ is not an object in the given category, but rather in the slice category, so that $A \to C$ belongs to the data of (what one also calls but should not ...) $A$. So actually $A \times_C B$ is an abbreviation of $(A \to C) \times (B \to C)$ (product in the slice category), where again the arrows should have names or a meaning clear from the context. $\endgroup$ Feb 24, 2014 at 20:17
  • 2
    $\begingroup$ In this case why don't we advocate to use the notation $f \times_C g$? That would make much more sense (in fact it will make sense at all), and is just as verbose as the (wrongish) notation $A \times_C B$. $\endgroup$
    – almaus
    Feb 25, 2014 at 7:38
  • 1
    $\begingroup$ Ok, I maybe understand. As $f \times_C g$ is, in a category with products, a subobject of $A \times B$, one does use the mongrel notation $A \times_C B$ that refers both to the subobject of $A \times B$ and to the product in slice category $\times_C$. I was heavily bathing in examples where there were no products (thus pullbacks were not their subobjects) so I lost this meaning. $\endgroup$
    – almaus
    Feb 25, 2014 at 7:52
  • 1
    $\begingroup$ @almaus, I agree, something like $f \times_C g$ would be better. Or perhaps $f_C \times g_C$, where $f_C$ denotes $f$ viewed as an object of the slice category into $C$. Or maybe even $C_f \times C_g$. $\endgroup$ Apr 27, 2014 at 15:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .