Lecture notes for Friday, Feb. 10

We already discussed the definition of the fundamental groupoid of a space and said it was an important (although simple) invariant of a topological space. But an invariant is much more useful if besides assigning a value to each object of interest, it also gives a way of comparing those values whenever you have a morphism between two objects. A simple way this can be arranged is to make your invariant into a functor, which we will now do for the fundamental groupoid.

The fundamental groupoid is a functor from the category of spaces to the category of groupoids. Let’s review what those categories are:

Spaces. The objects are spaces [1], and the morphisms are continuous functions.

Groupoids. The objects are groupoids and the morphisms are the things we called simply morphisms of groupoids; these are just functors between groupoids (recall that groupoids are just categories with all morphisms invertible).

We’ve already defined what the fundamental groupoid functor does on objects: to each space \(X\) it associates the groupoid \(\pi_{\le 1}X\) of homotopy (rel. endpoints!) classes of paths in \(X\). Now let’s say what it does on morphisms. Given a continuous function \(f : X \to Y\), we need to define a morphism of groupoids \(f_* : \pi_{\le 1} X \to \pi_{\le 1} Y\) (the name \(f_*\) is traditional). To do that, we need to say what \(f_*\) does to the objects and morphisms of \(\pi_{\le 1} X\), there is only one reasonable choice, namely:

1. An object \(x\) of \(\pi_{\le 1} X\) is just a point \(x\) of \(X\). We define \(f_*(x) = f(x)\), a point in \(Y\).
2. A morphism \(g : x_1 \to x_2\) in \(\pi_{\le 1}\) is a homotopy class of paths in \(X\) going from the point \(x_1\) to the point \(x_2\). We need to define \(f_*(g)\) as some homotopy class of paths in \(Y\) from \(f(x_1)\) to \(f(x_2)\). The obvious thing to do is to take any path \(\alpha\) in the equivalence class \(g\) and set \(f_*(g)\) to be the class of \(f \circ \alpha\), i.e., just apply \(f\) to every point on the path \(\alpha\). Using \([\ ]\) to denote “homotopy class (rel. endpoints) of”, we can write \(f_* ([ \alpha ]) = [f \circ \alpha]\).

Is this really a functor? We need to check the follwing things:

1. \(f_*\) is well-defined: if \(\alpha\) and \(\beta\) are homotopic paths from \(x_1\) to \(x_2\) in \(X\), then the paths \(f \circ \alpha\) and \(f \circ \beta\) are homotopic paths in \(Y\) (and go from \(f(x_1)\) to \(f(x_2)\)). This follows by just composing the homotopy between \(\alpha\) and \(\beta\) with \(f\).
2. \(f_*\) preserves identities: if \(\alpha\) is a constant path in \(X\), \(f \circ \alpha\) is constant too (we only needed that \(f \circ \alpha\) is homotopic to a constant, but the easiest way to show that is to observe it is constant).
3. \(f_*\) respects composition: if \(\alpha\) and \(\beta\) are two composable paths in \(X\), we need \(f \circ (\beta \alpha)\) to be homotopic to the composition \((f \circ \beta) (f \circ \alpha)\). Inspecting the definition shows these two paths are in fact equal, not just homotopic.

For now we won’t really do anything with the fact that we could extend the definition of \(\pi_{\le 1}\) on spaces to a functor, but this will be important later on.

OK, so we’ve improved our invariant of spaces to a whole functor defined on the category of spaces. Now let’s improve it in a different way: trim it down to a more manageable size! The fundamental groupoid of a space \(X\) has as objects all of the points of \(X\), of which most nice spaces have uncountably many. It’s usually more convenient to focus on a few points of interest, indeed, the traditional theory of the fundamental group says you can mostly get by with looking at a single object \(x_0\) (and all morphisms \(x_0 \to x_0\)) in the fundamental groupoid. That’s a little too drastic, lots of stuff is clearer using several basepoints at once. (This will be clearer once we talk about the van Kampen theorem and covering spaces.)

Here the definition: given a space \(X\) and a subset \(A\) of the points of \(X\), we define the fundamental groupoid of \(X\) with basepoints in \(A\) to be the groupoid \(\pi_{\le 1}(X,A)\) whose objects are the points of \(A\), and where the morphisms \(a_1 \to a_2\) are exactly what they are in \(\pi_{\le 1}X\), namely homotopy classes of paths in \(X\) from \(a_1\) to \(a_2\). Notice that the morphisms make use of the whole space \(X\), not jsut the subset \(A\).

In categorical terms, \(\pi_{\le 1}(X,A)\) is the full subcategory of \(\pi_{\le 1}X\) spanned by the objects in \(A\). Here “subcategory” indicates we are taking only some of the objects and morphisms, and the word “full” refers to the fact that we take all morphisms between objects that we include in the subcategory.

The fundamental group of a space \(X\) with basepoint \(x_0\) is then simply \(\pi_{\le 1}(X,\{x_0\})\). (Recall that a groupoid with a single object is really nothing but a group.) For this special case we typically write \(\pi_1(X,x_0)\).

turned into a discussion of representable functors

We will talk about free groupoids on directed graphs and groupoids given by a presentation next time, but first let’s review the simpler case [3] of free groups and presentations of groups.

The main points I want to make about free groups and group presentations are that:

1. They’re not hard to “understand”: you can easily write down elements of such a group, multiply them, etc.;
2. but the construction is a little bit fiddly,
3. it can be hard to prove things about them,
4. a good way to understand them is through the functors they represent.

Let’s start by looking at presentations of groups. A presentation of a group is given by specifying some generators for the group and some relations that these must satisfy. We write \(G = \langle g_1, \ldots \mid r_1 = \ldots = 1 \rangle\) to specify that \(G\) is generated by the \(g_i\) subject to the condition that the \(r_i = 1\) in \(G\). We allow the possibility of having zero relations or even zero generators. The group \(G\) presented in such a way satisfies:

• it has elements called \(g_1 , \ldots\), and therefore also the elements that the group laws force to exist, namely, all products of the form \(g_{i_1}^{e_1} g_{i_2}^{e_2} \cdots\), where the exponents are arbitrary integer (i.e., are allowed to be negative),
• different formulas of the form above can actually designate the same element of \(G\) but this only happens if it is possible to prove that the expressions are equal by using the axioms in the definition of group and the “extra axioms” that say that each \(r_i=1\) in \(G\).

This really does determine which group is meant by a presentation but in a roundabout way. Given a presentation you easily write down elements of the group and, say, multiply them. But given two expressions for elements in such a group it’s not always easy to figure out whether they represent the same element or not. In fact here isn’t always an algorithm that can do it. There is no algorithm that given a finite presentation and two words in the generators can tell you whether the words represent equal elements or not. Now, you might think that maybe the difficulty is just with making the algorithm work for arbitrary presentations, but that for any given presentation you might be able to give a specially tailored algorithm, or at least given a group, you might always be able to find a presentation for which there is such an algorithm. Well, no and still no! The existence of an algorithm that tells when words are equal doesn’t depend on the specific presentation chosen for a finitely presentable group, only on the isomorphism class of the group and furthermore, while for some (“easy”) groups there are algorithms for this, there exist specific groups for which it can be proven that no algorithm exists!

To read more about computability questions related to group presentations, take a look at MacTutor’s page on word problems for groups, which also gives plenty of historical detail. For a more in depth and technical survey, look at Chuck Miller’s Decision problems for groups: survey and reflections.

A further point I want to make about group presentations is that while the idea behind them is fairly simple and the description given above really is enough to characterize them, formally constructing them is a little fiddly. Take the simplest case: when there are no relations. A group given by a presentation of the form “this is a set of generators and there are no relations” is called the free group on the set of generators, and as a special case of what we said above, the group a) contains all products of powers (positive and negative) of the generator and b) two such products are equal only if it can be shown to follow from the group axioms. What if we really want to construct this group in a completely rigorous way? I guess we have two main approaches:

1. The free group is the set of equivalence classes of products of powers of the generators, where two words are equivalent if one can be obtained from the other by a sequence of steps where at each step you either cancel or insert a portion of a word of the form \(g g^{-1}\) or \(g^{-1} g\).
2. The free group is the set of reduced (i.e. “fully simplified”) products of powers of the generators, i.e., products where any two consecutive powers of generators appearing, \(g_i^{e_i}\) and \(g_j^{e_j}\) have distinct bases \(g_i\) and \(g_j\).

In both cases there are tedious checks to perform:

For approach 1 you need to check the product is well-defined. Basically you want to show something like “no matter what order you perform cancellations in, you get the same fully simplified word at the end”. For approach 2, you need to check associtivity, which boils down to pretty much the same thing as what’s needed in the previous approach. Now, none of this is all that hard, but it is annoying and, I would say, less than ideal for a definition of something as basic as a free group. I’d much rather have a very simple definition and then a proposition about the existence of free groups than a definition that requires proof!

To see how we’ll define free groups and groups given by presentations, let’s look at what makes free groups special: it’s very easy to define groups homomorphisms out of a free group, indeed, you can send the generators whereever you want! The basic philosophy of category theory is that, loosely speaking, it doesn’t matter what an object is, only what it does; more precisely, an object is determined up to isomorphism by its morphisms to all other objects. This means that the property we just mentioned can actually be the definition of free group:

Definition. A group \(G\) is free on a set \(X\) of generators, if for any group \(H\), every function \(X \to H\) extends uniquely to a group homomorphism \(G \to H\).

Another way to say this is as follows:

to give a group homomorphism from \(G\) to \(H\) is the same thing as to give a function from the set \(X\) to the set of elements of \(H\).

Now, this is a little vague (what is “the same thing as”?), and we could try to formalize it as:

A group \(G\) free on the set \(X\) if there is a bijection \({\text{Hom}}_{{\text{Groups}}}(G,H) \cong {\text{Hom}}_{{\text{Sets}}}(X,H)\) (where the \(H\) on the right is just the set of elements of \(H\)).

But this is missing something. What we really want to say is not just that there is some random bijection for each \(H\), but that they are all “defined in a similar way”. We could say, for example, consider the function \({\text{Hom}}_{{\text{Groups}}}(G,H) \to {\text{Hom}}_{{\text{Sets}}}(X,H)\) given by restricting an homomorphism (which is after all just a function) to \(X \subset G\); this specific function is a bijection.

It turns out to generalize better to other similar situations instead to say only that the bijection \({\text{Hom}}_{{\text{Groups}}}(G,H) \cong {\text{Hom}}_{{\text{Sets}}}(X,H)\) is natural in \(H\). This means there is some bijection chosen for each \(H\) and they are all “compatible” with morphisms between different \(H\)s. More precisely, this compatibility means that for every morphism \(f : H \to H'\) in the category of groups, we have functions \({\text{Hom}}_{{\text{Groups}}}(G,H) \to {\text{Hom}}_{Groups}(G,H')\) and \({\text{Hom}}_{{\text{Sets}}}(X,H) \to {\text{Hom}}_{Sets}(X,H')\) (both of these are given by composing with \(f\)) and that the following square commutes:

\({\text{Hom}}_{{\text{Groups}}}(G,H)\)	\(\to\)	\({\text{Hom}}_{Sets}(X,H)\)
\(\downarrow\)		\(\downarrow\)
\({\text{Hom}}_{{\text{Groups}}}(G,H')\)	\(\to\)	\({\text{Hom}}_{Sets}(X,H')\)

To say the square commutes means that the composition of the top and right arrows equals the composition of the left and bottom arrows; this is the “compatibility” with morphisms we had talked about earlier.

In general, this idea of a compatible family of morphisms between (the values taken by) two functors is captured by the concept of natural transformation:

Definition. Given two categories \(C\) and \(D\), and two functors \(F, G : C \to D\), a natural transformation \(\eta : F \Rightarrow G\) is a collection of morphisms in \(D\), one for each object \(X\) of \(C\), \(\eta_X : FX \to GX\), such that for any morphism \(f : X \to X'\) in \(C\), the following square commutes:

\(FX\)	\(\to\)	\(GX\)
\(\downarrow\)		\(\downarrow\)
\(FX'\)	\(\to\)	\(GX'\)

(Here the horizontal arrows are \(\eta_X\) and \(\eta_{X'}\) and the vertical arrows are \(Ff\) and \(Gf\).)

In this terminology, we are saying that \(G\) is the free group on a set \(X\) if there is a natural isomorphism (this just means a natural transformation all of whose compnents \(\eta_X\) are isomorphisms) between the two functors \({\text{Groups}}\to {\text{Sets}}\) [4], one given by \(H \mapsto {\text{Hom}}_{{\text{Groups}}}(G,H)\) and the other by \(H \mapsto {\text{Hom}}_{{\text{Sets}}}(X,H)\). Here I said “functors”, but only explained what the so-called functors do to objects in the category of groups. This ellision is common practice when the definition on morphisms is “obvious”; here, to be explicit, the first functor sends a morphism \(f : H \to H'\) to the function \({\text{Hom}}_{{\text{Groups}}}(G,H) \to {\text{Hom}}_{{\text{Groups}}}(G,H')\) given by \(h \mapsto f \circ h\), and similarly for the second functor.

In this last formulation, there is an operation that is left implicit: taking the set of elements of a group and thinking of it as just a set. If we want to make it explicit, it is natural to think of it as a functor [5] \({\text{Groups}}\to {\text{Sets}}\). We’ll call it \(U\) for underlying set functor; since it forgets about the group operation it is also called a forgetful functor. The operation of taking the free group on a set of generators is also a functor we will denote \(F : {\text{Sets}}\to {\text{Groups}}\). The relationship between them claimed above is that for any fixed set \(G\), we have a natural isomorphism of the following two functors \({\text{Groups}}\to {\text{Sets}}\): \(H \mapsto {\text{Hom}}(FX,\_)\) and \(H \mapsto {\text{Hom}}(X,UH)\). In fact more is true: the isomorphism is natural in both \(X\) and \(H\) –although it is contravariant in \(X\), which means that given a fixed \(H\) and a morphism \(X \to X'\) you can write down a naturality square like the one for a morphism \(H \to H'\) we saw above, except that the vertical arrows go up instead of down.

Whenever this situation arises, i.e., we have two functors \(F : C \to D\) and \(U : D \to C\) and a bijection \({\text{Hom}}_C(FX,Y) \cong {\text{Hom}}_D(X,UY)\) which is natural in \(Y\) and contravariantly natural in \(X\), we say that \(F\) and \(U\) are an adjoint pair of functors, \(F\) is the left adjoint of \(U\) and \(U\) is the right adjoint of \(F\).

Adjoint functors are extremely important in all of mathematics, an we’ll have more to say about them later. (In particular, at some point we’ll prove that left adjoints preserve pushouts –after we’ve defined pushouts, of course.)

[1] Recall out convention: space means topological space if you’re comfortable with those, or subset of \(\mathbb{R}^n\) if not.

[2] the last two points are the informal part.

[3] simpler mainly because you take the free group on a set of generators but you take the free groupoids generated by a directed graph.

[4] Warning: here the target category of these functors is \({\text{Sets}}\), not because we are talking about the free group on a set of generators, but just because the collection of all morphisms between two object in one of our categories is a set.

[5] We’re claiming here that it is a functor, which is obvious since group homomorphisms are special kinds of functions and their composition is just the composition as functions.