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?

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
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.


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
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 $\mu$.

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