Functoriality of fundamental group

Question

I'm trying to prove the following statement: If $\mathcal{C}_1$ is the category of path connected topological spaces and $\mathcal{C}_2$ is the category of groups, then the mapping $\mathcal{C}_1 \ni X \mapsto \pi_1(X)\in \mathcal{C}_2$ is not a functor. ($\pi_1(X)$ is the fundamental group of $X$).

Idea: I want to see that there is no way of associating to a morphism $f: X \rightarrow Y$ in $\mathcal{C}_1$ a morphism $f_{*}: \pi_1(X)\rightarrow \pi_1(Y)$ in $\mathcal{C}_2 : \forall f,g \in Mor(\mathcal{C}_1)$ that can be composed, having the equation: $$(g \circ f)_{*}= g_{*} \circ f_{*}$$

But I don't see how can I check that. I'd appreciate your help

Answer 1

$\def\ZZ{\mathbb{Z}}$I will show that such a functor cannot "look like" $\pi_1$. Let $S^1$ be the unit circle, let $C$ be the cylinder $S^1 \times [0,1]$ and let $K$ be the Klein bottle, the result of gluing the ends of $C$ together with the reverse orientation. We have $\pi_1(S^1) \cong \pi_1(C) \cong \ZZ$ and $\pi_1(K) \cong \ZZ \ltimes \ZZ$ where the action is by $-1$. We have the following maps:

• $\nu: S^1 \to S^1$, the map that reverses orientation.

• $i_1$ and $i_2: S^1 \to C$, the inclusions of the ends of $C$.

• $g: C \to K$, the map that glues the ends of the cylinder together.

Suppose $G$ were a functor from path connected spaces to abelian groups. If $G$ "looked like $\pi_1$", then we would expect

• $G(\nu) : \ZZ \to \ZZ$ should be $-1$.

• $G(i_1)$ and $G(i_2)$ should induce the same bijection $\ZZ \to \ZZ$.

• $G(g)$ should be an inclusion.

This is impossible. Notice that $g \circ i_1 = g \circ i_2 \circ \nu$, so the first two conditions imply that the generator of $G(g)$ sends the generator of $G(S^1)$ to a $2$-torsion element. But the only $2$-torsion element in $G(K)$ is the identity, contradicting the third condition.

I can prove that the first two conditions must hold. (More generally, I can prove that the functors $G^{\mathrm{ab}}(X) \otimes \mathbb{Q}$ and $H_1(X, \mathbb{Q})$ are naturally isomorphic.) But I am stumped trying to rule out the possibility that $G(g)$ is the trivial map. I don't want to work on this problem further, so I'm putting this up as a partial answer.

Answer 2

The problem is not in the morphisms … it is in the very definition of $\pi_1$ on objects: which group do you choose? There are, in principle, infinitely many groups $\pi_1(X,x)$ with $x\in X$, all different (because the underlying sets are different) but isomorphic. The isomorphism $\pi_1(X,x)\cong \pi_1(X,y)$ is not canonical, as it depends on the choice of a path from $x$ to $y$, so you have as many isomorphisms as many homotopy classes of paths. Therefore, there is no canonical way to pick a group in the group category and call it the fundamental group of $X$. So you cannot even define the map between objects.