The fundamental group of a pointed topological space $(X,T,x)$, is the group whose elements are homotopy equivalence classes of paths from $x$ to $x$, and whose composition is the concatenation of such paths.

This induces a function from the class of pointed topological spaces to the class of groups.

Moreover, every continuous function between topological spaces induces a homomorphism between their topological groups.

These two together form a functor $F:\textbf{Top}\to \textbf{Grp}$.

Why is it interesting that this is a functor? What can we do with this fact?


I guess the reason we care about functors in general is that a common way of studying an object $o$ in a category is by studying morphisms $\varphi$ going into/coming out of your object, and so by having a functor we not only get a (possibly simpler) object $F(o)$ but we ALSO get morphisms $F(\varphi)$ going into/coming out of it. Essentially it's preserving how all the objects are related.

Edit: As Joseph Martin points out in the comments to this answer, a functor is also in some sense the "correct" type of mapping between categories. Just like we only care about homomorphisms of groups rather than just plain functions, a functor is something which preserves the essential structure of the object it is being applied to, in this case a category. And just like we have a category $Grp$ where the morphisms are group homomorphisms, we also have $Cat$ where the objects are categories and the morphisms are functors. This is part of the reason why algebraic topologists take great pains in showing that their construction is functorial, because a plain "function" between the object classes forgets too much important structure.

One specific reason toplogists care about functors is that if $F\colon \mathcal{C} \to \mathcal{D}$ is a (covariant in this case) functor and $\varphi \in Mor_{\mathcal{C}}(x, y)$ is an isomophphism, then $F(\varphi) \in Mor_{\mathcal{D}}(F(x), F(y))$ is an isomorphism as well (I leave this as an exercise for you, and to show the analogous statement holds for contravariant functors).

Moreover $\pi_1$ is a homotopy functor, meaning that if $f\sim g$ then $\pi_1(f) = \pi_1(g)$, and hence $\pi_1$ defines a functor on the homotopy category $hTop$ whose objects are the same as $Top$ but whose morphisms are homotopy classes of continuous functions. In particular it follows that a homotopy equivalence of spaces induces an isomorphism of fundamental groups. Therefore if two spaces have non-isomorphic fundamental groups then it's actually functoriality which tells us they are not homotopy equivalent, for example $S^2$ and $T^2$.

Another application of functoriality is to show there is no retraction of the disk to the circle (a corollary of this fact is the Brouwer Fixed Point Theorem). Let $\iota \colon S^1 \to D^2$ denote the inclusion and suppose there is a continuous function $r \colon D^2 \to S^1$ such that $r\circ \iota = id_{S^1}$. Then by functoriality

$$ \pi_1(r) \circ \pi_1(\iota) = id_{\pi_1(S^1)} $$

But since $D^2$ is contractible $\pi_1(D^2) = 0$, so $id_{\pi_1(S^1)}$ must be $0$ since it factors through the $0$ group, which contradicts the fact that $\pi_1(S^1) \cong \mathbb{Z}$. Moreover if you know $\pi_n (S^n) \cong \mathbb{Z}$ then this argument also shows there is no retraction from $D^{n+1}$ to $S^n$.

The non-existence of a retraction could also have been proven using the functoriality of homology, after all $H_n(S^n)$ is much easier to compute. All we need for this argument is some functor $F$ such that $F(S^n)\neq 0$ and $F(\iota) \colon F(S^n) \to F(D^{n+1})$ is the $0$ morphism (so the target category of $F$ should at least have a zero object).

Another Edit: Tobias Kildetoft brought up in the comments a slightly more advanced application of functoriality: using the fact that $\pi_1$ preserves products, functoriality actually implies that $\pi_1(G)$ is abelian for any topological group $G$.

It's more advanced because it requires you know about Group Objects. The idea is that you can define a group as a set $G$ with some functions $m$, $inv$, and $e$ (coming from a one-point set) which satisfy some commutative diagrams involving $G$ and products of $G$. A "group object" in a category $\mathcal{C}$ with finite products is just an object $G$ of $\mathcal{C}$ and some morphisms $(m, inv, e)$ (where now $e$ comes from a terminal object) which make the same diagrams commute. In particular if we have a functor $F\colon \mathcal{C}\to \mathcal{D}$ which preserves products, functoriality implies that $F$ takes group objects in $\mathcal{C}$ to group objects in $\mathcal{D}$.

For example the group objects in $Set$ are groups, and if we add a topology and require the structure functions to be continuous then we see that topological groups are the group objects in $Top$. You might think that every group is a group object in $Grp$ but this is not the case. Remember that the structure on a group object requires morphisms, and for a non-ablian group $G$ the inversion function is not a group homomorphism; therefore if a non-abelian group admits the structure of a group object, it cannot be induced by its internal group operation and must be some external one. But, if you have a group $(G,\cdot)$ that admits a group object structure $(G, m, inv, e)$, then you can use an Eckmann-Hilton argument to show that $m = \cdot$ and in fact the operation is abelian. Indeed the group objects in $Grp$ are exactly the abelian groups, where the group object structure is induced by the internal group structure.

Now the argument that $\pi_1(G)$ is abeliean just goes like this: $G$ is a group object in $Top$, and $\pi_1$ preserves products, so by functoriality $\pi_1(G)$ is a group object in $Grp$, i.e. it's abelian.

A typical way of establishing this result more directly is to use the Eckmann-Hilton argument on the group $\pi_1(G)$, since it has two operations satisfying the interchange law (loop concatenation and point-wise composition). This argument using group objects has the advantage of pushing the formality into a more appropriate setting, so that now we know ANY functor $F\colon Top \to Grp$ that preserves products will take topological groups to abelian groups, and we get an analogous statement for functors between any other two categories with products by computing the group objects.

  • $\begingroup$ I assume $\pi_1$ is the functor that maps topological spaces to their fundamental group? $\endgroup$ – user56834 Mar 15 '19 at 14:11
  • $\begingroup$ Yes, by $\pi_1$ I mean the fundamental group functor. I have left out basepoints just to make things easier to look at. $\endgroup$ – William Mar 15 '19 at 14:12
  • 2
    $\begingroup$ I would also add that a functor is simply a 'map' between categories that 'preserves' the category structure (that is, it maps objects so that identities and composition are respected), so you can be sure that this relationship between categories that you want to use won't mess up the obvious things you care about. They are really no different to isometries, linear maps or even order in a poset. $\endgroup$ – Joseph Martin Mar 15 '19 at 14:23
  • 1
    $\begingroup$ Another consequence (though this also uses that the functor preserves products) is that the fundamental group of a topological group is abelian. $\endgroup$ – Tobias Kildetoft Mar 15 '19 at 19:53
  • 1
    $\begingroup$ I think I know what you mean. The idea is that since it preserves products it sends group-objects to group-objects, which in $Top$ are topological groups and in $Grp$ are actually abelian groups, by an Eckmann-Hilton argument. If that's right, it's a pretty cool "refactoring" to push the formal proof from the topological context into a purely $Grp$ context. Do you mind if I edit it into my answer? $\endgroup$ – William Mar 15 '19 at 20:16

If you only look at $\pi_1$ on objects, it's a very rough invariant, it's only a "discrete" invariant.

That it's a functor adds some distinguishing power, or more generally some power to the invariant.

First of all, functoriality implies that it's invariant under homeomorphism, because every functor preserves isomorphisms. But then you might argue that we only need the invariance under homeomorphism/homotopy equivalence, not the full functoriality.

But functoriality allows you to prove a bunch of stuff, like for instance that it preserves retracts, which is essentially what William uses to prove that $D^2$ doesn't retract onto $S^1$.

More generally, algebraic topology isn't interesting if you only look at spaces, and many paradigmatic problems of algebraic topology (lifting problems, extension problems, etc.) can be (or are ) phrased in terms of certain maps existing and certain maps composing to give other maps etc : functors preserve this structure, and so the functoriality of $\pi_1$ (more generally of functors you encounter in algebraic topology : $\pi_n$, $H_n, H^n$, etc.) allows you to really transform these topological problems in algebraic problems, which are often easier to study and understand.

Again, William's example is great with that : it starts with a topological problem ("is there a retraction of $D^2$ onto $S^1$ ?"), shows that a positive answer to this problem implies a positive answer to the corresponding algebraic problem ("if there were, there would be a retraction of $0$ on $\mathbb{Z}$") where we know that the algebraic problem actually has a negative answer ("there is no retraction of $0$ onto $\mathbb{Z}$"); and this power specifically comes from functoriality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.