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A groupoid is defined to be a category where every morphism is an isomorphism. So sometimes a group is said to just be a groupoid with one object.

When I try to make sense of this, I denote the single object as $G$. I view the morphisms as the analogue of "elements." The identity $1_G$ is the analogue of the usual identity $e$, we can compose any two morphisms since they are all arrows on $G$, and for any arrow $f$, we have some $f^{-1}$ such that $f\circ f^{-1}=f^{-1}\circ f=1_G$, so the idea of inverses is still there.

So I informally associate the elements of the group in the usual definition to be the arrows in the groupoid. But what does the sole object $G$ in the groupoid "correspond to" if I were to try to informally make sense of a group in the usual sense? Does it even correspond to anything?

Metapod
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    Often is doesn't even receive a name, but is simply denoted by $\bullet$, in latex code \bullet – Stefan Hamcke Nov 27 '13 at 15:13
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    It's worth noting that you don't need objects to define a category. You can view a category as a partial algebra on the class of morphisms, in which case a groupoid is just a partial (associative, unital) algebra in which all elements have inverses. The group is the special case where the groupoid operation is total. – Malice Vidrine Nov 30 '13 at 01:22

5 Answers5

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The single object in the groupoid corresponding to a group $G$ does not really correspond to anything in the group - but you can think of it as a thing which has as its group of symmetries the group $G$.

As an example, the category whose only object is the set $[n]=\{1,\dotsc,n\}$, and morphisms are bijections from this object onto itself. Then the symmetries of $[n]$ are given by the group $S_n$.

For a more subtle example, consider a topological space $X$, and fix a point $x\in X$. Consider the category with one object, namely the point $x$. A morphism from $x\to x$ is a homotopy-class of paths from $x$ to $x$. Composition of morphisms comes from concatenation of paths. The automorphism group of $x$ is the fundamental group $\pi_1(X,x)$, which is a group of symmetries of the based topological space $(X,x)$.

However, if you consider groupoids with many objects all of which are isomorphic to each other, then the different objects correspond to different realizations of the same group. Moreover, each isomorphism between different objects gives rise to an isomorphism between the corresponding realizations of the group.

To make this precise, suppose $\mathcal G$ is a groupoid where all objects are isomorphic. For each object $x$ of $\mathcal G$, let $G_x$ denote the group $\mathrm{Mor}_\mathcal G(x, x)$. A morphism $\phi:x\to y$ defines an isomorphism $G_y\to G_x$ by taking $g\mapsto \phi\circ g\circ \phi^{-1}$.

A nice example of this is the category $\mathrm{FB}_n$ whose objects are finite sets of cardinality $n$ and morphisms are bijections. Then the automorphism group of each object is isomorphic to $S_n$. Isomorphisms between objects determine isomorphisms of their automorphism groups.

Another example is the Fundamental Groupoid of a path connected topological space $X$. Its objects are the points of $X$. The set $\mathrm{Mor}(x,y)$ is the set of homotopy classes of paths from $x$ to $y$. In this groupoid, $G_x$ is the fundamental group $\pi_1(X,x)$ of $X$ based at the point $x$. Different base points result in isomorphic fundamental groups, and isomorphisms are determined by homotopy classes of paths between these points.

Amritanshu Prasad
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Disappointing as the answer may be, it doesn't really correspond to anything.

All the group axioms refer to elements of the group; and these elements correspond with the morphisms in the category. We don't care what the object is: there's only one object, after all! So all we're bothered about is the morphisms.

Clive Newstead
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  • I know this is an old question, but..... My primary issue is: Suppose we have two maps $f,g\in Hom(\bullet,\bullet)$ then necessarily $f(\bullet)=g(\bullet)$ and because $f$ and $g$ agree on all elements, namely $\bullet$, then why isn't it the case that $f=g$. Now clearly $f\neq g$ isn't true (in general), so is it wrong to think of $Hom$ as a collection of functions. In other words, in general there may be several maps from one element to another? – Squirtle Nov 01 '17 at 17:36
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    Here $\bullet$ is the domain and codomain of the morphism. The symbol $\bullet$ is just an arbitrary name given to the unique object of the group (considered as a category). There is no notion of 'element of $\bullet$' that it makes sense to ask whether $f$ and $g$ to agree on, and there is no reason in general to expect $f$ and $g$ are even functions. (Moreover, I could cook up concrete categories which have morphisms $f,g : X \to Y$ that agree on all elements of $X$ but are not equal as morphisms of the category.) – Clive Newstead Nov 01 '17 at 17:45
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    thank you, this makes sense. I'm new to studying category theory and thought that $Hom$ was a collection of functions, but I see that it is not. – Squirtle Nov 01 '17 at 20:21
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I think of a group as an abstraction of the invertible transformations of something. Since it is an abstraction, we don't worry about what that "something" is.

A one-object groupoid says exactly this, viz., that we are dealing with invertible transformations of "something". The single object of the groupoid is the "something".

Uday Reddy
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G doesn't correspond to anything. That is more or less by design; all of the structure of the group is in its morphisms, there isn't anything else to contribute.

That said, you can make it correspond to something. If $\mathcal{G}$ is the one-object groupoid, then you can consider functors $F : \mathcal{G} \to \mathcal{C}$ for various categories $\mathcal{C}$. This functor sends $G$ to an object $F(G)$ of $\mathcal{C}$; you can view this functor as selecting something for $G$ to correspond to.

For example, if $\mathcal{C} = \mathbf{Set}$, then more or less by definition, functors $F : \mathcal{G} \to \mathbf{Set}$ are the same thing as a left group action of the group $\hom(G,G)$ acting on the set $F(G)$.

Similarly, if $\mathcal{C} = \mathbf{Vect}_{\mathbb{C}}$, then functors give representations of the group $\hom(G,G)$ acting on complex vector spaces, if $\mathcal{C} =\mathbf{AbGrp}$ we get left $\hom(G,G)$-modules, and so forth.

It is not terrible to interpret $G$ as a sort of universal "thing upon which $\hom(G,G)$ acts".

Alternatively, in the categorical approach to formal logic, we might consider the internal language of $\mathcal{G}$, and interpret $G$ as being the group, and each arrow in $\hom(G,G)$ gives a (generalized) element of $G$.

This interpretation is closely related to applying the Yoneda embedding to $\mathcal{G}$ and to the Cayley representation of the group $\hom(G,G)$ acting on itself.

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I think it is best to step back and comprehend the following more general statement:

a monoid as a category with one object

Because upon adding invertibility a monoid becomes a group and a category becomes a groupoid and your statement follows.

As to why a monoid can be viewed as a category with one object Chilango's answer to this question should be clear.

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