Lecture notes for Friday, Feb. 3

Introducing Groupoids

The problem set for this week is to do the exercises contained below, these are also listed on their own here

Introducing Groupoids

A small change.

The main goal for today will be to present the definition of groupoids and several examples. The main example for this tutorial is the fundamental groupoid of a space, that will be left for the next meeting.

Two definitions of groupoid

There is a definition of groupoid that is very similar to the definition of group, intuitively “a groupoid is like a group with a partially defined multiplication”. The precise details are as follows:

First definition of groupoid. A groupoid is a set with a partially defined multiplication such that

\(a(bc) = (ab)c\), that is, whenever one side is defined so if the other and they are equal; additionally, when both \(ab\) and \(bc\) are defined, so is \(a(bc)\);
every element \(a\) has an inverse \(a^{-1}\) such that both \(aa^{-1}\) and \(a^{-1}a\) are defined, these products behave as identities “as much as possible”, namely, whenever \(ab\) is defined, \(a^{-1}ab=b\) and \(abb^{-1}=a\).

Exercise. Show that if in a groupoid both \(ab\) and \(ac\) are defined \(b^{-1}b = c^{-1}c\).

Exercise. Show that in a groupoid \((a^{-1})^{-1} = a\).

Notice that given a set \(X\), which sets of pairs of elements of \(X\) can be the domain of definition of a groupoid multiplication is only specified very indirectly by the axioms: they say things like “if the domain of definition include such and such pair it must also contain this other pair”, etc. Think of multiplication of matrices or of composition of functions. For those operations the domain of definition is more explicitly specified: two matrices \(A\) and \(B\) can be multiplied if the number of rows of \(A\) equals the number of columns of \(B\); two functions \(f\) and \(g\) can be composed when the domain of \(f\) equals the codomain of \(g\). It turns out that the domain of the multiplication of any groupoid can be described in that way, that is, for any groupoid there is a notion of source and target of an element so that \(ab\) is defined if and only if the target of \(b\) equals the source of \(a\). Once these sources and targets are given we are very close to defining a category, so let’s just do that:

Second definition of groupoid. A groupoid is a category where all the morphisms are invertible, and,

Definition of category. A category \(\mathcal{C}\) consists of

a collection of objects,
for every pair of objects \(X\) and \(Y\), a collection of morphisms with source \(X\) and target \(Y\), and
a composition operation \(g \circ f\) on morphisms that is defined for any pair of morphisms \(f\) and \(g\) where \(f : X \to Y\) and \(g : Y \to Z\).

So just like a groupoid is a group but with a partially defined composition, a category is a monoid but with a partially defined composition, i.e., a category is what a “monoidoid” would be if that were a word.

Exercise. Prove that the two definitions are equivalent. This is easy but a little long and was sketched in class. First of all, let’s figure out what we mean by this equivalence: it means that any groupoid according to the first definition can be regarded as the collection of morphisms of a category with invertible morphisms, and that given any category with all morphisms inverible, it’s morphisms (more precisely, the disjointified union of its morphism collections \(\hom_C(X,Y)\)) form a groupoid according to the first definition. The second statement is clear, to prove the first you just need to do the following: given a groupoid according to the first definition, define for it objects, and sources and targets of elements and prove that \(ab\) is defined if and only if \(\text{source}(a) = \text{target}(b)\). (Very briefly explain why this is enough.) We saw two methods for doing this:

Method 1. Define \(\text{source}(a) = \{ b : ab\ \text{is defined}\}\), and \(\text{target}(a) = \text{source}(a^{-1})\).

Method 2. Define the objects to be the collection of identities \(\{aa^{-1}\}\) and then \(\text{source}(a) = a^{-1} a\), \(\text{target}(a) = a a^{-1}\).

A short ramble about category theory

Most kinds of structures studied in mathematics form categories: people usually don’t just define objects, they also define special maps between them which form the morphisms of a category. Examples of this kind of categories are (listed as “objects and morphisms”):

sets and functions,
groups and group homomorphisms,
Abelian groups and group homomorphisms,
vector spaces over a given field \(K\) and linear transformations,
topological spaces and continuous functions,
smooth manifolds and differentiable maps (if you don’t know what a manifold is, here is a subexample: open subsets of \(\mathbb{R}^n\) –with varying \(n\), say– and differentiable functions),
partially ordered sets and order preserving (or monotone) functions,
finite graphs and adjacency preserving maps between vertex sets (usually called graph homomorphisms), etc.

A first level of familiarity with category theory is recognizing that lots and lots of objects studied in mathematics are the objects of some category; a second level is realizing that many objects are themselves special kinds of categories. Here we will only mention some very simple examples:

Since a group is a special case of a groupoid (when the multiplication is everywhere defined) and a groupoid is a special case of a category, a group is also a special kind of category. Unwinding the definitions, a group is a category that only has one object and all of whose morphisms are invertible. (Notice that when a category only has one object every pair of morphisms can be composed.)
As above, but without requiring inverses: a monoid is a category with only object.
A partial order is almost category such that for every pair of objects \(X\) and \(Y\) there are either no morphisms \(X \to Y\), or only one.

Exercise. Explain why the word “almost” appears in the previous example and show how to fix the definition of a partial order as a special kind of category.

Remark. To make these definitions of familiar objects as special kinds of categories meaningfull, one should check that the preexisting notions of morphisms between them are equivalent to just functors between the categories we are attempting to redefine them as.

Examples of groupoids

Any group is a groupoid.
More generally, given any collection of groups \(G_1\), \(G_2\), …, their disjoint union \(G = G_1 \sqcup G_2 \sqcup \cdots\) is a groupoid; here a pair of morphisms of \(G\) can only be composed if they come from the same \(G_i\) in which case their composition is the product they have there.
A set can be regarded as a groupoid that only has identity morphisms; these are called discrete groupoids.
Given a category \(\mathcal{C}\) if we keep all the objects but only the invertible morphisms we get the core of \(\mathcal{C}\).
(One I forgot to mention in class.) An indiscrete groupoid is one where there is a unique morphism between any pair of objects, i.e., it is a category where all the objects are isomorphic to each other in a unique way.
(Another one I forgot) A disjoint union of indiscrete groupoids is basically an equivalence relation.
Given a group \(G\) acting on a set \(X\), we define the action groupoid of the action to have as objects the elements of \(X\), and as morphisms¹ between \(x, y\) in \(X\), all elements of \(G\) such that \(g \cdot x = y\) –where composition is just multiplication in \(G\). The action groupoid is denoted by \(X//G\).

Exercise. What is the action groupoid of the action of \(G\) on itself by left multiplication?

The fundamental groupoid of a space! This will be the main example of groupoid for us, but I ran out of time today before introducing it.

Things briefly hinted at to be covered in more detail later

Functors

Functors are the good notion of morphisms between categories, and in particular between groupoids. We sped through the definition:

Definition. A functor \(F\) between two categories \(\mathcal{C}\) and \(\mathcal{D}\) associates to each object \(X\) of \(\mathcal{C}\) an object \(FX\) of \(\mathcal{D}\) and to each morphism \(f:X \to Y\), a morphism \(F(f):FX \to FY\) in such a way that 1) \(F(\text{id}_X) = \text{id}_{FX}\) and 2) \(F(g \circ f) = F(g) \circ F(f)\).

We mentioned as examples that if you regard groups as single-object categories,

a functor \(G \to H\) is nothing but a group homomorphism, and
a functor \(G \to \text{Set}\) is a set \(X\) together with an action of \(G\) –we even said without explanation that the Grothendieck construction of this functor is just the action groupoid of the action.

Equivalence of categories, skeleta and a “classification” of groupoids

The skeleton of a category is formed by picking one object out of each isomorphism class of objects (and taking all existing morphisms between the chosen objects). Two categories are equivalent if they have isomorphic skeleta, or in other words, two equivalent categories only differ in the numbers of (redundant) isomorphic copies they have of each object.

This gives a sort of classification of groupoids: each groupoid is equivalent to a disjoint union of groups. Some people feel that this implies that groupoids are not worth studying on their own, since their study reduces to group theory. I hope this tutorial will convince you otherwise. And in the meantime, I will say that this argument against groupoids breaks down for structured groupoids, for example, if say, the set of objects has a topology.

Notice here that these morphism sets are not disjoint from one another, so making this groupoid into a first-definition-groupoid requires disjointifying them. For example, we could say the elements of the groupoid are triples \((x,g,y)\) such that \(g\cdot x = y\), with multiplication \((x,g,y)(y,h,z) = (x,hg,z)\) defined by the grace of the matching \(y\).