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Let $A\rightrightarrows X$ be a groupoid, where $X$ is the set of objects and $A$ is the set of arrows. My favorite example of a groupoid is an action groupoid. If a group $G$ acts on the left on a set $X$, we set $$ A=\{(x,g,y)\mid x,y\in X, g\in G,\ y=g*x\}, $$ then $A\rightrightarrows X$ with the evident maps is called the action groupoid corresponding to the action of $G$ on $X$. It is often denoted by $G\ltimes X$.

Let $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ be a morphism of groupoids (a functor). We say that $F$ is an equivalence of groupoids if it is an equivalence of categories.

Let $x\in X$. We denote by $A(x)$ the set of arrows $a\colon x\to x$. We denote by $\pi_0(X)$ the set of connected components of $X$ (i.e., the set of equivalence classes in $X$ with respect to the equivalence relation induced by $A$). We say that a morphism $F$ as above is a weak equivalence of groupoids (or a quasi-isomorphism) if $\pi_0(F)\colon \pi_0(X)\to \pi_0(Y)$ is a bijection and, for any $x\in X$, the induced homomorphism $F_x\colon A(x)\to B(y)$ is an isomorphism, where $y=F(x)$.

Question 1. Is it true that any weak equivalence of groupoids is an equivalence?

Now assume that a group $\Gamma$ acts on our groupoid $A\rightrightarrows X$. We say that $A\rightrightarrows X$ is a $\Gamma$-groupoid. My favorite example of a $\Gamma$-groupoid comes from an action of an algebraic group $\mathcal{G}$, defined over a field $k$, on a $k$-variety $\mathcal{X}$. Let $k_s$ denote a separable closure of $k$, then we set $\Gamma:={\rm Gal}(k_s/k)$, and we consider the action groupoid $\mathcal{G}(k_s)\ltimes\mathcal{X}(k_s)$, on which $\Gamma$ acts.

By a weak equivalence of $\Gamma$-groupoids we mean a $\Gamma$-functor $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ that is a weak equivalence of groupoids. By an equivalence of $\Gamma$-groupoids we mean a $\Gamma$-functor $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ such that there exists a a $\Gamma$-functor $F'$ in the opposite direction and each of the composite functors $F\circ F'$ and $F'\circ F$ is $\Gamma$-naturally-isomorphic to the corresponding identity functor.

Question 2. Is it true that any weak equivalence of $\Gamma$-groupoids is an equivalence of $\Gamma$-groupoids?

I expect the answer "No" to Question 2, but I cannot construct a counter-example.

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The braces that I typed in the displayed formula are not visible! Maybe somebody can edit the formula to make the braces visible... Thank you, –  Mikhail Borovoi May 15 '13 at 13:05
    
@Mikhail Just use \left\\{ instead of \left\{ and \\{ instead of \{ etc. (Demo: $\left\\{\right\\}$ vs. $\{\}$) –  The User May 15 '13 at 13:35
    
In comments it works differently—strange. –  The User May 15 '13 at 13:35
    
@The User: Thank you for the answer and the editing suggestion! –  Mikhail Borovoi May 15 '13 at 13:52
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1 Answer

up vote 5 down vote accepted

The answer to question 1 is “yes” since every weak equivalence is essentially surjective (let $y\in Y$, $[y]$ the connected component of $y$, which is in the image of $\pi_0(F)$, thus there exists $y^\prime\in [y]$ which is an element of the image of $F$ and isomorphic to $y$) and full and faithful (since $F_x$ is an isomorphism, the mappings $F_{xy}\colon A(x,y)\to B(F(x),F(y))$ are also bijective, choose $f\in A(y,x)$ then $\alpha\colon A(x,y)\to A(x), g\mapsto fg$ and $\beta\colon B(F(x),F(y))\to B(F(x)), g\mapsto F(f)g$ are bijections and $F_{xy}=\beta^{-1}\circ F_x\circ \alpha$ is bijective).

As a counter example for question 2 consider: $\Gamma=C_2$, $X$ the “complete” (in the sense of graph theory) groupoid consisting of two objects and four isomorphisms. The action of $\Gamma$ on $X$ is defined to swap the objects. Let $Y$ be the terminal groupoid consisting of a single isomorphism. $\Gamma$ acts trivially on $Y$. The unique morphism $X\to Y$ is a weak equivalence of $\Gamma$-groupoids. But there does not exist a $\Gamma$-morphism $Y\to X$.

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