I'm new to category theory and trying to convince myself that the definition of a product makes sense. Ultimately my question is: in what sense does the category theory definition of a product generalize the usual definition? What is the intuition? The rest is an explanation of what I've tried so far:

In an introduction I was looking at (ch. 7), it uses the notion of a well-pointed category to convert talk about elements to talk about arrows. So a product of $X$ and $Y$ is an object $O$ and arrows $\pi_1: O \rightarrow X$ and $\pi_2: O \rightarrow Y$ such that for any two arrows $\overrightarrow{x}: 1 \rightarrow X$ and $\overrightarrow{y}: 1 \rightarrow Y$ for a terminal object $1$ (in Set, just a singleton), there is a unique $\overrightarrow{u}: 1 \rightarrow O$ such that this diagram commutes:

enter image description here

So I understand this as for any pair of elements $x \in X$ and $y \in Y$, there is a unique $u \in O$ that projects on to both of them. The arrows pick out the elements. That makes sense to me. Then we jump to the general definition, which works in any category: $[O, \pi_1, \pi_2]$ is a product of $X$ and $Y$ iff for every $S$, $f_1:S \rightarrow X$, and $f_2:S \rightarrow Y$, there is a unique $u:S\rightarrow O$ such that this diagram commutes:

enter image description here

I don't really understand why we can just do that. I think it would be illuminating to see a proof that if $[O, \pi_1, \pi_2]$ is a product in the well-pointed sense, then it must be a product in the general sense. I've attempted it, but I keep getting stuck. I keep staring at this diagram:

enter image description here

I'm trying to convince myself that a $u$ that makes the diagram commute exists and is unique if I assume the rest of the diagram commutes, but I feel like I'm missing something. Any help is appreciated. Thanks.

  • $\begingroup$ Well, you know where $u$ must send every point of $S$, right? $\endgroup$ – Kevin Carlson Apr 5 at 2:40
  • $\begingroup$ @KevinCarlson If it were just a set and the usual ordered pairs, make $u(s) = (f_1(s), f_2(s))$. But I feel like I don't know the internal structure of $O$ in a general category. Or am I just going in circles for no reason? $\endgroup$ – Cordello Apr 5 at 2:45
  • $\begingroup$ Just write $\pi_i \circ u\circ s=f_i\circ s$, which uniquely determines $u\circ s$. Unfortunately as Malice says this only determines $u$ under a stronger form of well pointedness which is extremely rare. $\endgroup$ – Kevin Carlson Apr 5 at 5:31

The text is not claiming that the universal property of the second diagram is a consequence of the first; such an assertion is not generally true, even in the well-pointed case (see here for a counterexample).

It is rather expressing a more general definition of a product, so there is nothing to be proved here. Smith himself alludes to the fact that a definition in terms of points doesn't give you much in, for instance, the category $\mathbf{Grp}$ (every group is a product of any two groups in that sense). But the direct product of groups does satisfy the latter, more general form in non-trivial ways.

In case it helps to get the intuition, let's look at the category of sets, where the property expressed in the first diagram does imply the more general property. In $\mathbf{Set}$ every $x\in S$ corresponds to a unique point (once we pick a particular terminal object) $p_x:1\to S$. In $\mathbf{Set}$, any $S$-indexed family of points $i_x:1\to T$ will induce a function $i:S\to T$ with $i\circ p_x=i_x$ (this is a much stronger condition than Smith supposes, that $1$ is a dense generator). So in particular, any $S$-indexed family $u_x:1\to A\times B$ will give rise to a function $u:S\to A\times B$, but also two other $S$-indexed families $$\pi_1u_x:1\to A,\:\pi_2u_x:1\to B,$$ that then correspond to two maps $a:S\to A,\:b:S\to B$ with $ap_x=\pi_1u_x$ and $bp_x=\pi_2u_x$. In this special case you can check that $a,b$ are uniquely determined by the $u_x$ and vice versa.

But again, the situation I've described for $\mathbf{Set}$ is a very rare one. In general, you can't hope to infer that $O$ has the universal property in the second sense from it having it in the special case of maps from $1$.

  • $\begingroup$ Thank you, this helps. It also explains why I always felt like I was missing an arrow. In order to have it, I would need $1$ to be a dense generator. $\endgroup$ – Cordello Apr 5 at 13:56

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