# Explicit construction of the quotient of a category by a group action

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.

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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 $\mathcal C$. –  Matt Brin Jan 23 '12 at 18:48
$f$ 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. –  Roman Bruckner Jan 24 '12 at 10:06

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'.

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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.]

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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. –  Ryan Reich Jan 25 '12 at 0:20
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). –  Chris Schommer-Pries Jan 25 '12 at 15:07
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. –  Kevin Walker Jan 25 '12 at 15:47
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. –  Chris Schommer-Pries Jan 25 '12 at 17:27
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. –  Kevin Walker Jan 25 '12 at 18:06
There is also a nice answer in the case of a group $G$ acting on a groupoid $\Gamma$.
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.)