Note that I changed the theme again.
I’ve been spending a lot of time on the basic definition of the homotopy groups. Basically, in summary, we have defined for a pointed space , the group
of pointed homotopy classes . Alternatively, this is the group of pointed homotopy classes of maps ; this is because is homeomorphic to the sphere. This is a functor on the category of pointed topological spaces. If , then there is an induced homomorphism .
That’s a short summary of what I’ve discussed in these past few weeks; I won’t keep repeating it.
But now we want to fit these groups into exact sequences. This can be done directly in an ad hoc manner. Another way is to use the Puppe sequence.
1. General nonsense
Recall that a sequence of pointed sets
(with maps preserving the basepoints) is called exact if
In other words, the “kernel” at one end (i.e. the set mapping into the basepoint of ) is the image of .
In defining exact sequences in a category, we use this notion.
So fix a category with a zero object . This means that is simultaneously an initial and a terminal object. It follows easily from this that for any two , there is a unique which factors through . Namely, it is
This is called the zero morphism. In particular, the hom functors take values in the category of pointed sets.
Definition 1 Let be a category with zero object. A sequence
in the category is called exact if for each , the sequence
is exact as a sequence of pointed sets. The sequence is said to be coexact if the sequence
is exact as a sequence of pointed sets for all objects .
I don’t think this definition is a perfect recapitulation of the usual definition of exactness in an additive category. In an additive category, (even the category of abelian groups) it is generally not true that exactness of implies that of . So this is something a little different; it’s more analogous to split exactness in standard additive/abelian category theory.
Exact sequences in
The reason we would want to think this way is evident. Just as there is an exact sequence of homology, there is an exact sequence of homotopy groups. (To do this, we will have to define the relative homotopy groups—more on this shortly.) Since the homotopy groups are obtained by homming in the homotopy category of pointed topological spaces, it makes sense to talk about exact and coexact things in this category.
First, we have to note that has a zero object. Namely, this is the one-point space (whose basepoint is the unique point!). So the zero morphism between two spaces is the map which sends everything to the basepoint. That’s why the hom-sets are pointed.
So let be a basepoint-preserving morphism of pointed spaces . We want to construct a space with a sequence of pointed spaces
which is coexact in . This means that the composition is nullhomotopic and a map out of extends over if and only if its pullback to is nullhomotopic.
Here is the construction of . Namely, we start with the reduced cone of : this is the smash product , or alternatively
This is a pointed space into which embeds; moreover, the inclusion is nullhomotopic (relative to the basepoint!). We just shrink the entire cone to the top edge, which has been pinched off into a point.
We let . This is the space obtained from the disjoint union by attaching the points to . Then is a pointed space, and there is a canonical inclusion which preserves basepoints.
Proposition 2 The sequence
is coexact in .
Proof: Indeed, the composite is nullhomotopic because we can consider the family of embeddings
As goes to , this goes into the basepoint of .
Conversely, suppose we have a map nullhomotopy of . Then we can use and together to define a map on into . So if is nullhomotopic, then it extends to . This proves coexactness.
The Barratt-Puppe sequence
Now we want to extend this coexact sequence further. Well, the natural thing to do would be to write (with denoting the inclusion ) a sequence:
This is a natural extension of the exact sequence we gave above. However, it’s also rather messy. Fortunately, it turns out that these terms are homotopy equivalent to simpler things.
Here is a drawing of one of the spaces.
Let’s start with the term . As the figure shows, this is basically the union of two cones such that the end of the cone is glued into part of the end of . The “small cone” is glued into the “big cone” . Now, the big cone is contractible. So, intuitively, we should be able to quotient by it and get something homotopy equivalent. When we quotient by the big cone, then we are left with , except that its base has been identified to a point. So we get , which is just the reduced suspension discussed earlier.
Proposition 3 The space is homotopy equivalent to .
This follows from the above quotienting argument, if we grant the assumption that we are allowed to quotient by the contractible guy . I will come back to why this is allowed later.
We can play the same game for the seemingly more complicated term
and argue that we can quotient by the cone . In this case, we are just left with the suspension . Moreover, the map obtained from the map by collapsing these two terms is homotopic to the suspension of , i.e. . This takes a little self-convincing, and is best left to the reader.
So we find that our coexact sequence becomes
Here the first map is the map that collapses and to a point. The second map is the suspension of .
Proposition 4 Let . Then there is a coexact sequence in :
Recall that a long sequence of elements in is coexact if every three-term sequence is. This is analog with how normally you define exactness for additive categories. Every three terms of this is, up to homotopy equivalence in , of the form in some manner (with varying). So we find that this five-term sequence is exact.
Theorem 5 (Barratt-Puppe) Given , there is a long coexact sequence in , going .
Proof: Indeed, we know that the first five terms of this form a coexact sequence. This is just the previous proposition. In general, any three consecutive terms in this sequence are the suspension of a three-term subsequence of the first five terms. But any three consecutive terms in the first five are coexact. So the result follows from the next lemma:
Lemma 6 Suppose is coexact in . Then the suspension is coexact as well.
Proof: This follows purely formally because has a right adjoint . Indeed, let be a space; we show that
is exact as a sequence of sets. However, this sequence is just
which is exact by coexactness of . This proves the lemma.
So we have shown the theorem. The associated sequence is called the Barratt-Puppe or simply Puppe sequence associated to the morphism .
The loose end
Technically, it remains to explain why we had the homotopy equivalence
In other words, we should explain why we were allowed to quotient by the contractible space and not change things up to homotopy equivalence. For the explanation, I follow Bredon.
The reason is that there is a deformation of the space into itself carrying into a point. Formally, let us define this idea.
Definition 7 Let be a pointed space and a closed subspace containing the basepoint. A deformation of is a homotopy of the identity. That is, it is a continuous map such that and . We say that carries into a point if and .
Now, let us prove:
Proposition 8 Suppose is a pointed space and a closed subspace containing . Suppose there is a deformation carrying into . Then the quotient map
is a homotopy equivalence (homotopies preserving the basepoint).
Proof: Indeed, we define the map
via . Note that collapses to a point so factors through . I claim that are homotopy inverses.
First, consider . This is the map . It is homotopic to . So this is homotopic to the identity.
Next, consider . This is the map . But factors through because , so it becomes a map
In particular, this map is a homotopy between and .
So to finish the details in the proof of the Barratt-Puppe sequence, we just need to check that there is a deformation of that takes into a point. The idea is to push down into its “nose” while stretching slightly. I leave the details to the reader.