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Stream: learning: questions

Topic: Object that generate equivalent slice cats

Xuanrui Qi (Apr 28 2023 at 07:23):

So, given a category $\mathcal{C}$ and two objects $X, Y \in \mathcal{C}$, say we know that $\mathcal{C}/X$ and $\mathcal{C}/Y$ are equivalent, is there anything in particular that we can say about $X$ and $Y$? The answer seems like "no" to me, but can we extract even a little bit of information?

Morgan Rogers (he/him) (Apr 28 2023 at 07:49):

With no restrictions on the category, I don't think so. If you take any object in any category whatsoever, and consider the two copies of that object in the disjoint union of two copies of the category, the slices over the objects are isomorphic.

Joe Moeller (Apr 28 2023 at 13:22):

Here's a refinement of the question: instead of asking about the slices being equivalent, ask what you get when the functor between them given by postcomposition by some map gives an equivalence.

John Baez (Apr 28 2023 at 16:49):

Does that imply the map must be an isomorphism?

Kevin Arlin (Apr 28 2023 at 17:32):

Yeah! If $f$ makes an equivalence $\mathcal C/X\to \mathcal C/Y,$ then first note that by fully faithfulness, since the identity of $X$ is terminal in the domain, $f$ has only one endomorphism in the codomain. Now if $g:X'\to X$ is such that $f\circ g$ is isomorphic to the identity of $Y$ in $\mathcal C/Y$, then by inverting this isomorphism and composing the inverse with $g$ you get $h:Y\to X$ in $\mathcal C$ which is a mono, split by $f.$ Now, $f$ is a morphism $f\to \mathrm{id}_Y$ in $\mathcal C/Y,$ while $h$ is a morphism $\mathrm{id}_Y\to f,$ so composing we get an endomorphism of $f,$ which we started out by saying has to be the identity.

A couple of interesting things about this are to note that I've only really used fullness of the functor induced by $f,$ and to look on the other side of the Grothendieck construction. The slice categories are the Grothendieck constructions of the representable functors on $X$ and $Y,$ and generally the Grothendieck construction gives an equivalence between presheaves on $\mathcal C$ and discrete fibrations over $\mathcal C.$ So this argument shows that even though it seems like there can be equivalences of discrete fibrations while there can only be isomorphisms of presheaves, there's no real looser notion in this case. I'm not quite sure what's the right thing to say about the 2-category of discrete fibrations in general, but I guess it probably must be essentially discrete on 2-morphisms by a generalization of this argument, to be consistent with the Grothendieck construction result.