Given a finite group $G$, and a finite category $\mathcal{C}$, one can define the action of $G$ on $\mathcal{C}$ as a functor $A_{\mathcal{C}}\colon G\to\mathbf{Cat}$, which takes the single object of $G$ (regarded as a category) to $\mathcal{C}$. Moreover, one can define the quotient $\mathcal{C}/G$ to be the colimit of $A_{\mathcal{C}}$. There is an explicit construction of the category $\mathcal{C}/G$: let's denote the orbit of an element $a$ by $Ga$. The object set of $\mathcal{C}/G$ is simply given by the orbits of the elements of $\mathcal{C}^{(0)}$. To construct the set of morphism ${\mathcal{C}/G}^{(1)}$, one defines a relation $\leftrightarrow$ on $\mathcal{C}^{(1)}$ by saying $f\leftrightarrow g$ iff there is are decompositions $f = f_1\circ\...\circ f_n$ and $g=g_1\circ\...\circ g_n$, such that $G f_i = G g_i$ for all $i=1,...,n$. This relation is clearly symmetric and reflexive. It is however, not transitive. So one defines $\sim$ to be the transitive closure of $\leftrightarrow$ and sets $\mathcal{C}/G^{(1)} := \mathcal{C}^{(1)}/\sim$. My problem is that even though I can imagine why transitivity fails (i.e. given 3 morphisms $f,g,h$, one might find decompostions such that $f\sim g$ and $g\sim h$, but no decompositions such that $f\sim h$), I can't find an explicit example to demonstrate that case.

P.S.: As far as I know this might very well work too without the restrictions to finite groups and categories. But I haven't thought this through yet. So I restricted my question to the finite case.

  • 1
    $\begingroup$ What do you mean by $Gf_i=Gg_i$? If you mean there is some $h_i$ in $G$ so that $h_i(f_i)$ (regarding $h$ as a functor from $\mathcal C$ to itself), and $h_i$ is allowed to vary with $i$, then it seems to me that your relation is transitive. It all comes down to the definition of the action of $G$ on the category. My reading is that each element of $G$ becomes a self functor on $\mathcal C$. $\endgroup$
    – Matt Brin
    Jan 23, 2012 at 18:48
  • $\begingroup$ $f$ and $g$ were meant to be morphisms of $\mathcal{C}$, i.e. $f,g\in\mathcal{C}^{(1)}$. Though I admit the usage of g was a bit misleading. Then by $Gf_i=Gg_i$ I meant that $f_i$ and $g_i$ are in the same orbit: $G f_i = \{A_{\mathcal{C}}(x)(f_i) : x\in G\} = \{A_{\mathcal{C}}(x)(g_i) : x\in G\} = G g_i$ This construction is necessary to ensure, that composition of equivalenceclasses is well-defined. $\endgroup$ Jan 24, 2012 at 10:06

3 Answers 3


This question has been bugging me since it was posted, in part because I keep thinking you should have a homotopy action and should take the homotopy quotient.

But anyway.... whew! I have an example which demonstrates why the relation is not transitive. The example consists of a category C with four objects $x,y, y'$, and $z$. The group in question is the cyclic group of order two $G = \mathbb{Z}/2 \mathbb{Z}$ and the action on C fixes the objects identically.

The category C is almost a free category. Here are the generating morphisms:

  • morphisms $a_i$ for $i = 0,1$ which go from $x$ to $y$,
  • morphisms $a_i'$ for $i = 0,1$ which go from $x$ to $y'$,
  • morphisms $b_i$ for $i = 0,1$ which go from $y$ to $z$, and
  • morphisms $b_i'$ for $i = 0,1$ which go from $y'$ to $z$.

The effect of $k \in G$ on $a_i$, $a_i'$, $b_i$, or $b'_i$ is to send $i$ to $i+k$ mod 2. There are two relations which we impose to define this category:

  • $ b_0 a_0 = b_0' a_0'$, and
  • $ b_1 a_1 = b_1' a_1'$.

Notice that this pair of identifications is compatible with the group action. This means there are exactly six morphisms from x to z. Now all of them are equivalent under the relation generated by $\leftrightarrow$, but this relation is not transitive. For example we have:

$$b_1 a_0 \leftrightarrow b_1 a_1 = b_1' a_1' \leftrightarrow b_1' a_0'$$

but there is no direct relation between $b_1 a_0$ and $b_1' a_0'$.

A similar construction works with G any group with the index $i \in G$ given by group elements. In this case the quotient category, as you defined it, is the `free walking commutative square'.


This is more of a comment than an answer.

I think a nicer way to construct the quotient $\def\cC{\mathcal{C}} \cC/G$ is as follows. The objects of $\cC/G$ are the same as the objects of $\cC$. The morphisms of $\cC/G$ are the the morphisms of $\cC$ plus, for each pair $(x,p)$ in $G\times\cC^0$, a morphism $m_{x,p}: p\to x\cdot p$, where $x\cdot p$ denotes the result of acting on the object $p$ with the group element $x$. These new morphisms are required to satisfy the following obvious relations: $$ m_{x,p} \bullet m_{y,xp} = m_{xy,p}$$ and $$ f \bullet m_{x,p} = m_{x,q} \bullet (x\cdot f) .$$ [EDIT: and $m_{1,x} = id_x$.] This is for all $x,y\in G$ and all morphisms $f:q\to p$. I'm using the arrow convention for composition in the category: $f\bullet g$ instead of $g\circ f$, if $f:p\to q$ and $g:q\to r$.

In other words, we have added new (iso)morphisms relating $p$ to $x\cdot p$, and these new morphisms interact with the old morphisms according to the group action. If one skeletonizes this category by choosing a single object in each $G$ orbit, then one arrives at the description in your question. But this construction feels more natural to me.

[EDIT: Oops -- I was too hasty in claiming they are the same; see comments below. I'm pretty sure this is the homotopy quotient, but I'll stop sort of saying I'm certain since I haven't had time to think it completely through and I've already hit my being-wrong-in-public quota for this month.]

  • $\begingroup$ I was about to advocate this myself, so instead you get my vote. In the example of a sheaf of groups acting non-freely on a sheaf of sets, the book (for example) "Champs Algebriques" by Laumon and Moret-Bailly describes the construction of the quotient (as a stack) of an algebraic space by a sheaf of groups in this way. $\endgroup$
    – Ryan Reich
    Jan 25, 2012 at 0:20
  • $\begingroup$ Are you sure this is right? Consider the example where C is the group $\mathbb{Z}/3$ thought of as a one object category with $G = \mathbb{Z}/2$ acting on C by fixing the unique object and as the automorphisms of $\mathbb{Z}/3$. Then the presentation you give looks to me like it gives the dihedral group of order 6 thought of as a one object category, while Roman's quotient gives the terminal singleton category. (I'm assuming you also meant to add the relation $m_{1,p} = id_p$ otherwise your construction is even bigger). $\endgroup$ Jan 25, 2012 at 15:07
  • $\begingroup$ I think you're right -- there's something wrong with my claim. I'll think about it a little bit more before editing the answer. $\endgroup$ Jan 25, 2012 at 15:47
  • $\begingroup$ It looks to me like you are describing the homotopy quotient, which is something like $C \times_G EG$ where $EG$ is the pair groupoid on $G$. In contrast Roman's quotient is the strict quotient, which can be quite destructive. $\endgroup$ Jan 25, 2012 at 17:27
  • $\begingroup$ Yes, that's what I was about to comment but you beat me to it. A simpler (than you're Z/3 counterexample) counterexample (to my claim that the constructions are the same) is $G$ acting on the trivial category. $\endgroup$ Jan 25, 2012 at 18:06

There is also a nice answer in the case of a group $G$ acting on a groupoid $\Gamma$.

Taylor, J. Quotients of groupoids by the action of a group. Math. Proc. Camb. Phil. Soc. 103 (1988) 239--249.

One forms the composite $\Gamma \to \Gamma \rtimes G \to (\Gamma \rtimes G)/N$, where the second groupoid is the semidirect product and the first morphism is $\gamma \mapsto (\gamma,1)$; $N$ is the normal closure of all elements of the form $(0_x,g)$ for all $x \in Ob(\Gamma), g \in G$; and the second morphism is the quotient morphism. The theorem is that this composition is an orbit morphism, i.e. gives the quotient of $\Gamma$ by the action of $G$. (This normal closure is not usually a disjoint union of object groups.)

This result is applied to calculate the fundamental groupoid of certain orbit spaces in Chapter 11 of my book "Topology and groupoids" (2006). For example it works for discontinuous actions of groups on a Hausdorff space which admits a universal cover (joint work with Philip Higgins, also in the 1988 differently titled edition of that book). For example it calculates the fundamental group of the symmetric square of a space. (A version of this work is in arXiv:math/0212271.)


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