The Hurewicz Theorem

by ayushkhaitan3437

Here we talk about the Hurewicz theorem. Let X be a path connected space with \pi_n(X)=0 for n\geq 2. Then X is determined up to homotopy by \pi_1(X).

What does “determined up to homotopy” mean? It means that all the spaces that satisfy the condition above are homotopic to each other. When two spaces are homotopic, what does that mean? Say A and B are homotopic. This means that there exist maps f:A\to B and g:B\to A such that g\circ f\simeq id_A and f\circ g\simeq id_B. Can we think of any examples of spaces that are not homotopic? Yes. Just map a disconnected space to a connected space. Like mapping two non-intersecting discs to a connected disc. A homotopy between spaces tends to preserve the same “kind” of connectivity between spaces; i.e. two disconnected discs would be homotopic to two discs, and not three disconnected discs.

Why do we need the Hurewicz theorem? Because it is often difficult to calculate the fundamental group of a space, but much easier to calculate the homology groups. Hence, knowing that we can map homology groups to homotopy groups, we can discover properties of the homotopy groups that we couldn’t earlier.

We state and prove the Hurewicz Theorem for the n=1 case. Let \phi:\pi_1(X,x_0)\to H_1(X) be defined such that a path \gamma goes to \gamma. How is \gamma a member of the homology group? Because it is a map from [0,1], which is a 1-simplex, to X! Now for the map to be well-defined, if \gamma'\sim\gamma, then \phi(\gamma')=\phi(\gamma). Why is this true? We shall find out later.

Anyway, we have a homomorphism \phi:\pi_1(X,x_0)\to H_1(X), and H_1(X) is an abelian group. Hence, we have a homomorphism \phi':(\pi_1)_{ab}(X,x_0)\to H_1(X), where (\pi_1)_{ab}(X,x_0) is just the group \pi_1(X,x_0) modulo its commutator subgroup. We have abelianized the fundamental homotopy group. Hurewicz Theorem says that the map h' is an isomorphism. How do we see this? Moreover, we have not even proven that h' is a homomorphism, or even a well-defined map.

We need to prove three things: that h' is well defined, a homomorphism, and an isomorphism.

!. Well-defined- Consider two homotopic paths \gamma\sim\gamma'. We need to prove that \gamma and \gamma' belong to the same homology group. There exists a map H(s.t):I\times I\to X such that H(0,t)=\gamma(t) and H(1,t)=\gamma'(t). The solid square I\times I can be thought of as the union of two 2-simplices \sigma_1 and \sigma_2. We’ll orient the boundary edges in a compatible fashion, and consider the restriction of H to \partial \sigma_1 and \partial \sigma_2.

On a completely different but related topic, note that the constant map from \Delta^1\to X is just the boundary of the constant map from \Delta^2\to X. Coming back to the above argument, we note that H(\partial\sigma_1-\partial\sigma_2)=\gamma'-\gamma-2f_{x_0}. Here f_{x_0} just denotes the mapping of a 1-simplex to the point x_0.

So what exactly is happening here? How do we know that \gamma-\gamma'\in \text{Im}(H_2(X))? Because \gamma-\gamma'=H(\partial\sigma_1)-H(\partial\sigma_2)+2f_{x_0}, and all three of H(\partial\sigma_1), H(\partial\sigma_2) and 2f_{x_0} belong to Im(H_2(X). Hence, we’ve proved that \phi is well-defined.

2. Homomorphism: Let [\gamma] and [\delta] be elements of the fundamental group. Consider \gamma*\delta:I\to X. Moreover, let \Delta^2 be [v_0 v_2 v_3] and let \sigma:\Delta^2\to X. On the [v_0v_2] edge, let the restriction of \sigma be \gamma*\delta. On the other two edges, the restriction of \sigma should be \gamma and \delta.

First we need to prove that reparametrization does not affect homotopy. Why’s that? The homotopy lies in continuously deforming one interval to another, and then making the path (say f) act on it. This homotopy is clear. Now the proof says that \gamma-\gamma*\delta+\delta is a reparametrization of \gamma*\delta. How is that?

Anyway, we have h([\gamma])+h([\delta])-\partial\sigma=\gamma*\delta=h([\gamma]*[\delta]). Hence, we get a homomorphism.

3. Surjectivity: Let us take a 1-cycle \sigma=\sum_i\sigma_i. Here it is possible that \sigma_i=\sigma_j for i\neq j. We can re-write this sum as a sum of loops, but putting together non-loops in the obvious way. Hence, we know that all cycles can be mapped to. These cycles may be disjoint too. It’s just that we’ve covered all cycles.

Now let us move to more general 1-simplices. Let \gamma_i be the path from x_0 to the base point of \sigma_i. Then \gamma_i\sigma_i\overline{\gamma_i} is homologous to \gamma_i+\sigma_i+\overline{\gamma_i}. What does this mean? It means that their difference is a boundary. Why are they homologous? This is because h([\gamma_i][\sigma_i][\gamma'_i])=h([\gamma_i])+h([\sigma_i])+h([\gamma'_i]), as h is a homomorphism. Now \overline{\gamma_i} is homologous to -\gamma_i. Hence, we get \sigma_i is homologous to h([\gamma_i][\sigma_i][\gamma'_i]). By doing this, we can center all our loops at x_0. Now any sum of loops centred at x_0 can be mapped to from the fundamental group, as that is what the fundamental group is, the collection of all loops centred at any point (remember that X is path-connected). Hence, we’ve proven that \phi is surjective.

4. Kernel: We need to prove that the kernel is the commutator subgroup. We can see why \gamma\gamma'\overline{\gamma}\overline{\gamma'} belongs to the kernel. This is because h([\gamma][\gamma'][\overline{\gamma}][\overline{\gamma'}]) is homologous to \gamma+\gamma'-\gamma-\gamma'. Now we need to prove that the kernel is a subset of the commutator subgroup.

I’m not typing up the inclusion of the kernel in the commutator subgroup, but it can be found here.