Quotient of a category by a free group action 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?
 A: 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.
A: 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.
A: 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.
