Here we talk about the Hurewicz theorem. Let be a path connected space with
for
. Then
is determined up to homotopy by
.
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 and
are homotopic. This means that there exist maps
and
such that
and
. 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 case. Let
be defined such that a path
goes to
. How is
a member of the homology group? Because it is a map from
, which is a
-simplex, to
! Now for the map to be well-defined, if
, then
. Why is this true? We shall find out later.
Anyway, we have a homomorphism , and
is an abelian group. Hence, we have a homomorphism
, where
is just the group
modulo its commutator subgroup. We have abelianized the fundamental homotopy group. Hurewicz Theorem says that the map
is an isomorphism. How do we see this? Moreover, we have not even proven that
is a homomorphism, or even a well-defined map.
We need to prove three things: that is well defined, a homomorphism, and an isomorphism.
!. Well-defined- Consider two homotopic paths . We need to prove that
and
belong to the same homology group. There exists a map
such that
and
. The solid square
can be thought of as the union of two
-simplices
and
. We’ll orient the boundary edges in a compatible fashion, and consider the restriction of
to
and
.
On a completely different but related topic, note that the constant map from is just the boundary of the constant map from
. Coming back to the above argument, we note that
. Here
just denotes the mapping of a
-simplex to the point
.
So what exactly is happening here? How do we know that ? Because
, and all three of
and
belong to
. Hence, we’ve proved that
is well-defined.
2. Homomorphism: Let and
be elements of the fundamental group. Consider
. Moreover, let
be
and let
. On the
edge, let the restriction of
be
. On the other two edges, the restriction of
should be
and
.
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 ) act on it. This homotopy is clear. Now the proof says that
is a reparametrization of
. How is that?
Anyway, we have . Hence, we get a homomorphism.
3. Surjectivity: Let us take a -cycle
. Here it is possible that
for
. 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 -simplices. Let
be the path from
to the base point of
. Then
is homologous to
. What does this mean? It means that their difference is a boundary. Why are they homologous? This is because
, as
is a homomorphism. Now
is homologous to
. Hence, we get
is homologous to
. By doing this, we can center all our loops at
. Now any sum of loops centred at
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
is path-connected). Hence, we’ve proven that
is surjective.
4. Kernel: We need to prove that the kernel is the commutator subgroup. We can see why belongs to the kernel. This is because
is homologous to
. 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.