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A groupoid is a category in which all morphisms are invertible.(*) The groupoids form a very nice subclass of categories. The inclusion of the groupoids into the 2-category of small categories admits both left and right (weak) adjoints. So you can localize (or complete) a category to a groupoid. If E denotes the free walking isomorphism, then the groupoids are precisely the categories which are local with respect to the inclusion of the free walking arrow (let's denote this as [1]) into E.

I have been thinking a lot recently about a certain very nice class of categories, which in many ways is the antithesis of the groupoids. These are the categories X, such that a morphism is an isomorphism in X if and only if it is an identity. Let us call these categories lean.(**) These too form a nice class of categories. The inclusion admits a left adjoint, and the lean categories are precisely those categories which are local for the projection from [1] to the terminal category pt.

What is this left adjoint? (We'll denote it as L). It forms a certain quotient category. You can see it explicitly by taking your given category and forming the quotient where you identify all isomorphisms with identities.

In some cases it is easy to work out what you get. For example if you have a groupoid X and you apply L, you get a discrete category (set) consisting of the components of X.

More generally you have to be more careful. Quotienting out by just the isomorphisms is not compatible with the composition in your category, and so the result is not a category. To get the actual categorical quotient you have to quotient by additional things: compositional consequences of the initial equivalence relation.

For example suppose that f,g,h are morphisms in your category X, where g and fgh are isomorphisms. Then for any lean category Y and any functor $X \to Y$, both g and fgh must map to identities in Y. Hence the composite fh must also map to an identity, even if it is not an isomorphism of X. It is not hard to construct an example where this happens. Nevertheless we can ask the following

Question: Suppose that X is a category such that the lean quotient L(X) is a set (discrete category). Is X necessarily a groupoid?

If true, I think this would give a very neat characterization of the groupoids. I have so far been unable to prove this result or come up with a counter example, but I feel I have been staring at the problem too long. It is enough to prove just the case that L(X) is just a single point. The above discussion shows that in general the quotient $X \to L(X)$ identifies things which are not isomorphisms, which suggests the answer is negative. A positive answer would have to somehow make use of the fact that everything is being identified.

(*) For this question when I speak of categories I mean small categories.

(**) If anyone knows of a reference for these categories, I would be interested to know it.

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I was thinking about these recently, and I convinced myself that there is no right adjoint to the inclusion of lean categories into $Cat$. –  David Roberts Sep 13 '11 at 23:46
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3 Answers 3

up vote 10 down vote accepted

I believe the following is an example of a category whose leanification is discrete but which is not a groupoid. It should be possible to simplify it. There may be details to work out. Let $B$ be the bicyclic monoid. It is generated by $a,b$ subject to the relations $ab=1$. (Note $ba\neq 1$). Let $K(B)$ be the Karoubi envelope of $B$. The objects of $K(B)$ are the idempotents of $B$, so $b^na^n$ with $n\geq 0$. An arrow is a triple $(e,m,f)\colon e\to f$ where $fme=m$. The product is $(f,n,g)(e,m,f)=(e,nm,g)$.

I claim $L(K(B))$ is the trivial category (1 object with only the identity morphism). First of all, it is easy to see that any two objects of $K(B)$ are isomorphic. Namely, $(b^ma^m,a^m,1)$ is an isomorphism of $b^ma^m$ with $1$ with inverse $(1,b^m,b^ma^m)$. So in $L(K(B))$ there is only one object. It is well known that every proper quotient of $B$ is a group. So it suffices to show the endomorphism monoid at $1$ collapses. Since $(1,b,ba)$ is the identity in $L(K(B))$ we observe that $(1,ba,1)=(ba,ba,1)(1,b,ba)(1,a,1)$ which is then equivalent to $(1,ba^2,1)$ in $L(K(B))$. Thus the endomorphism monoid at $1$ collapses to a group and hence to a point since automorphisms are trivial in $L(K(B))$. The same happens then to all other endomorphism monoids (as all objects isomorphics).

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You beat me by seconds... –  Todd Trimble Sep 13 '11 at 22:29
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I think the answer is negative. For this I'm going to adapt a passage from Categories for the Working Mathematician (bottom paragraph of page 164 in the 2nd edition). The counterexample involves a skeleton of the category of countably infinite sets, which I'll denote by $C$; let $\mathbb{N}$ be the unique object.

For a morphism $f$ in $C$, let $[f]$ be its image under the quotient $C \to L(C)$. We have

$$[f \times (g \times h)] = [(f \times g) \times h]$$

given that the associativity is identified with the identity on $\mathbb{N} \times \mathbb{N} \times \mathbb{N} = \mathbb{N}$. Letting $p_1, p_2$ be the two projections from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$, we have

$$[f p_1] = [p_1(f \times (g \times h))] = [p_1] [f \times (g \times h)] = [p_1] [(f \times g) \times h]$$

$$= [p_1 ((f \times g) \times h)] = [(f \times g) p_1]$$

where the first and last equations follow from naturality of projections in $C$. Hence

$$[f] [p_1] = [f \times g] [p_1]$$

in $L(C)$. However, $[p_1]$ is epi in $L(C)$; indeed, every epi in $C$ splits assuming the countable axiom of choice, and any functor preserves split epis. Hence $[f] = [f \times g]$. A similar argument with $p_2$ in place of $p_1$ gives $[f \times g] = [g]$; hence $[f] = [g]$ for any two morphisms $f, g$ in $C$. Thus $L(C)$ consists of only one object and one morphism, the identity.

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I got luck there. I think our examples are pretty much of the same spirit. –  Benjamin Steinberg Sep 13 '11 at 22:34
    
I wish I could accept both answers. They're great, thanks! –  Chris Schommer-Pries Sep 14 '11 at 11:17
    
You don't need the countable axiom of choice to split epis in $C$, since you're just choosing natural numbers and so you can choose the smallest available one. –  Andreas Blass Sep 14 '11 at 13:08
    
Thanks, Andreas -- you're right. I was writing quickly and that was the first thing that had occurred to me. –  Todd Trimble Sep 14 '11 at 14:39
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Here's a different (as far as I can see) proof for the same example Todd gave, the monoid of endofunctions of $\mathbb N$. Consider the following elements of this monoid:

  • $b$ is the bijection that interchanges $2n$ with $2n+1$ for all $n$.

  • $e:n\mapsto 2n$ and $f:n:\mapsto 2n+1$.

  • $p$ maps even numbers by $2n\mapsto n$ and maps all odd numbers to 0$.

  • $z$ is the constant zero map.

Then we have

  • $f=b\circ e$,

  • $p\circ e=1$, and $p\circ f=z$.

In the quotient monoid where all isomorphisms are identified, $b$ becomes 1, so the first of these equations gives $f=e$, and then the remaining equations give $1=p\circ e=p\circ f=z$. But then, for any $x$ in the monoid, we have in the quotient that $x=1\circ x=z\circ x=z$.

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This is a nice proof. –  Chris Schommer-Pries Sep 14 '11 at 13:41
    
I was looking for a proof like this last night, but you found it first! –  Benjamin Steinberg Sep 14 '11 at 13:58
    
Very neat, Andreas. –  Todd Trimble Sep 14 '11 at 14:38
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