10
$\begingroup$

Let's say we have a locally $\lambda$-presentable category and a pair of $\lambda$-presentable objects $A$ and $B$. Is it true that $A \times B$ is $\lambda$-presentable?

$\endgroup$
  • 1
    $\begingroup$ I don't think it's true, but I don't have a counterexample to hand. However, it is not true that the terminal object must be $\lambda$-presentable. For example, take $\mathbf{Set}^I$ where $I$ is any set; this is locally finitely presentable, but the terminal object is finitely presentable if and only if $I$ is a finite set. $\endgroup$ – Zhen Lin Dec 9 '14 at 10:28
  • $\begingroup$ On the other hand, it's not hard to show that in a locally presentable category, there are arbitrarily large regular cardinals $\mu$ for which the $\mu$-presentable objects are closed under binary product. $\endgroup$ – Tim Campion Jul 29 '18 at 16:15
10
$\begingroup$

In $\mathbf{Grp}$, the finitely presentable objects are precisely the finitely presented groups. Let $F_2$ be the free group on two elements. Then $F_2 \times F_2$ is finitely generated but not finitely presented, so the class of finitely presentable objects in $\mathbf{Grp}$ is not closed under binary products.

$\endgroup$
  • $\begingroup$ Nice dig there, Zhen Lin. $\endgroup$ – Todd Trimble Dec 9 '14 at 14:28
  • 1
    $\begingroup$ I'm probably just confused, but I don't see how this works. The linked question is not about the whole group $F_2\times F_2$ but a certain subgroup. Why isn't $F_2\times F_2$ presented by taking the four obvious generators and taking four relations saying that each generator of the first $F_2$ commutes with each generator of the second. $\endgroup$ – Andreas Blass Mar 12 at 18:27
6
$\begingroup$

A simpler counterexample is given by the slice category $S/\mathrm{Set}$ for a "large" (= of cardinality to be chosen later) set $S$. This category can be viewed as the category of models of an algebraic theory with one nullary operation for each element of $S$, so it is locally finitely presentable. Its initial object $S$ (equipped with the identity map) is of course finitely presentable, but $S \times S$ is a free object on the "large" set of pairs $\{\,(a, b) \mid a \in S, b \in S, a \ne b\,\}$. By choosing the cardinality $S$ large enough, we can then make $S \times S$ not only not finitely presentable, but not $\mu$-presentable for any fixed regular cardinal $\mu$.

$\endgroup$
  • $\begingroup$ Welcome back! Is it too early to get excited that you might start contributing again? $\endgroup$ – David White Aug 30 '18 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.