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?
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asked Dec 9, 2014 at 10:17
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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 LinCommented Dec 9, 2014 at 10:28
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$\begingroup$ 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 CampionCommented Jul 29, 2018 at 16:15
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2 Answers
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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.
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8$\begingroup$ 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. $\endgroup$ Commented Mar 12, 2019 at 18:27
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$\begingroup$ As noted above: the direct product of finitely presented groups is obviously a finitely presented group (combine generators and relations and add commutation relations of generators), so the category of groups is not a counterexample. $\endgroup$ Commented Nov 18, 2022 at 23:56
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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 $\mu$.
answered Aug 1, 2018 at 20:22
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1$\begingroup$ Welcome back! Is it too early to get excited that you might start contributing again? $\endgroup$ Commented Aug 30, 2018 at 12:54