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Stream: learning: questions

Topic: Is the category of pseudo-squares locally presentable?

Kevin Carlson (May 07 2025 at 01:24):

Bourke has done a great study of 2-categories that have a property you might call being 'pseudo-locally finitely presentable': they're finitely accessible, with flexible limits commuting with filtered colimits, and their retract equivalences are accessible. Bourke calls such 2-categories $LP_{\mathcal M}.$ A lot of his argument hinges on showing carefully that the power of an object $\mathcal K$ of $LP_{\mathcal M}$ by the generic arrow with respect to the pseudo-Gray structure, so the 2-category of arrows and pseudo squares in $\mathcal K,$ remains in $LP_{\mathcal M}.$

Kevin Carlson (May 07 2025 at 01:24):

In particular, this shows that the 2-category of functors and pseudo-squares, $\mathcal{PS}=\mathbf{Ps}(\to,\mathsf{Cat})$ is pseudo-locally presentable. But it seems like, in this case, it's actually strict locally presentable. Since it's accessible with products, this just comes down to the existence of equalizers, and I'm pretty sure $\mathcal{PS}$ has equalizers: the trick is they're not pointwise, in that in the below situation, you have to define $E_0$ as the intersection of (1) $\mathrm{eq}(f_0,g_0)$, (2) $a^*\mathrm{eq}(f_1,g_1)$, and (3) the equifier of $\varphi,\gamma.$ But with that unusual tweak it looks like a perfectly good equalizer ot me.
Screenshot 2025-05-06 at 6.22.28 PM.png

Kevin Carlson (May 07 2025 at 01:26):

Am I missing something, or am I right that pseudo-squares are locally presentable? If not, is there some general reason why, say, 2-categories of presheaf type in the pseudo sense should be locally presentable, or is this just a brute fact?

Vít Jelínek (May 11 2025 at 01:55):

@Kevin Carlson
Hmm, if I am not missing something, then this is not a correct description of equalizers. In principle, if you have another morphism $h:C\to A$, then there is no reason for $h_0$ to equify $\gamma$ and $\varphi$; therefore, you don't necessarily get an arrow in the proposed equalizer.

Vít Jelínek (May 11 2025 at 01:55):

To give a concrete counterexample, set $A_0=B_0=C_0=C_1=1$ (the terminal category) and $A_1=B_1=\mathbb{Z}$ (the category with one object and integers as morphisms). So the only data with some freedom for their definition is: $\eta$ (the 2-cell part of $h$), $\varphi, \gamma, f_1, g_1$. Moreover, all of those are defined by just choosing one integer. So if we set $\eta = 1, \varphi=1, \gamma=0, f_1=2, g_1=3$, we really get that $fh=gh$. However, since $E_0$ is empty, there cannot be any morphism $C\to E$, so $E$ can't be the equalizer.

Kevin Carlson (May 11 2025 at 02:49):

Oh, yes, I see, very interesting! A bad intuition around weighted limits. But isn’t your $C$ then the equalizer, at least in this case? At least your choice of $\eta$ looks to me to be forced.

Vít Jelínek (May 11 2025 at 11:17):

Yes, I believe that in this specific case, it is the equalizer, but I don't see how to generalize it unfortunately