Let Cat denote the 1-category of small categories. The functor Mor : Cat -> Set which assigns to a category its set of morphisms (aka Hom([• -> •], -)) does not commute with most colimits. Does it commute with quotients by free group actions? In other words, if C is a small category and G is a group acting on C such that the action of G on the objects of C is free, does Mor(C/G) = (Mor C)/G?
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asked Oct 15, 2009 at 19:19
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3 Answers
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The answer is yes, provided your action has the following data that you didn't specify:
- For any g in G and any f: a -> b, an assignment gf: ga -> gb, in a way that strictly respects multiplication in G (including 1f = f).
If you don't say what G does on morphisms, I don't see how the question makes sense.
answered Oct 15, 2009 at 22:48
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$\begingroup$ You're right, I meant to say that G acts on C as an object of Cat. Hopefully the new wording is better. That was the answer I expected, but why is it true? $\endgroup$ Commented Oct 15, 2009 at 22:55
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$\begingroup$ I guess the short answer would be that the set of all morphisms is a disjoint union of morphism sets between pairs of objects (this is often an explicit axiom in the definition of category), so the freeness of the G-action on objects translates to a free action on morphisms. I may be missing something - are there subtleties in your definition of quotient category C/G? $\endgroup$– S. Carnahan ♦Commented Oct 15, 2009 at 23:28
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$\begingroup$ I was defining C/G as the colimit of the diagram BG -> Cat given by the G-object C (where BG is the groupoid with one object which has automorphism group G). Usually when forming a colimit in Cat we have to throw in some compositions of morphisms (e.g. consider the quotient [• -> •] / [• ∐ •] which has infinitely many morphisms). But now you've convinced me that in my case I can just write down a category with objects (Ob C)/G and morphisms (Mor C)/G and check the universal property (functors out of it correspond to G-invariant functors from C). $\endgroup$ Commented Oct 16, 2009 at 0:10
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OK, I found a "high-tech" argument which is maybe more convincing.
The external input to the argument is that if X -> Z <- Y is a diagram of G-sets with G acting freely on Z (and hence on X and Y) then (X xZ Y)/G = (X/G) x(Z/G) (Y/G). This is pretty clear because we can always write our original X -> Z <- Y in the form (X/G) x G -> (Z/G) x G <- (Y/G) x G.
Now we use the presentation of Cat as the localization of the category of simplicial sets in which the local objects are the simplicial sets X such that Xn -> X1 xX0 ... xX0 X1 is an isomorphism for all n >= 2. By general nonsense about localizations, if C is a category (viewed as a simplicial set) with a G-action, we can compute the colimit C/G by taking the colimit levelwise and then localizing the resulting object. But the levelwise quotient is already local, by our previous observation. Since C1 represents the set of morphisms of C, we're done.
answered Oct 16, 2009 at 6:20
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This came up in something I'm working on, and this question was the first hit on google.
It seems to me that this has nothing to do with G acting freely, and that the answer is always yes. One can always define a quotient category C/G with objects (Ob C)/G and morphisms (Mor C)/G, with composition and identities defined in a more-or-less obvious manner (use the action of G to match up two arrows $f: C\to D$ and $h: gD\to E$ whose domains are equal in C/G, and then compose; note that $h\circ g(f)$ and $g^{-1} (h) \circ f$ are equivalent morphisms in the quotient). I was a little surprised that this makes sense, but it seems to.
I think it is elementary to check that C/G satisfies the universal property defining the colimit of the functor $BG \to$ Cat.
Am I missing something? Edit: YES! See the comments below. If you care to see what I was actually thinking about (a version of this which is actually correct) look at my answer to this subsequent question.
answered Apr 27, 2010 at 4:42
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$\begingroup$ What if I take C = BH, where H is a group with an action of G? I think the quotient category will be B(H with the relations x = gx for all x in H imposed), whose set of morphisms will be different from the set of orbits for the action of G on H as a set. $\endgroup$ Commented Apr 27, 2010 at 4:52
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3$\begingroup$ If the action is not free on objects, you have choices when picking the $g$ which makes the morphisms "match up". $\endgroup$ Commented Apr 27, 2010 at 4:52
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1$\begingroup$ Ah, I see. I think this saved me from a fair amount of time-wasting. I'll leave the answer up, as a warning... $\endgroup$ Commented Apr 27, 2010 at 5:40