The Barratt-Puppe sequence

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 ${(X, x_0)}$ , the group

$\displaystyle \pi_n(X, x_0)$

of pointed homotopy classes ${(S^n, s_0) \rightarrow (X, x_0)}$ . Alternatively, this is the group of pointed homotopy classes of maps ${f: (I^n, \partial I^n) \rightarrow (X, x_0)}$ ; this is because ${I^n/\partial I^n}$ is homeomorphic to the sphere. This is a functor on the category of pointed topological spaces. If ${f: (X, x_0) \rightarrow (Y, y_0)}$ , then there is an induced homomorphism ${f_*: \pi_n(X, x_0) \rightarrow \pi_n(Y, y_0)}$ .

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

$\displaystyle (A, a_0) \stackrel{f}{\rightarrow} (B, b_0) \stackrel{g}{\rightarrow} (C, c_0)$

(with maps preserving the basepoints) is called exact if

$\displaystyle f(A) = g^{-1}(c_0).$

In other words, the “kernel” at one end (i.e. the set mapping into the basepoint of ${C}$ ) is the image of ${A}$ .

In defining exact sequences in a category, we use this notion.

So fix a category ${\mathfrak{C}}$ with a zero object ${\ast}$ . This means that ${\ast}$ is simultaneously an initial and a terminal object. It follows easily from this that for any two ${X, Y \in \mathfrak{C}}$ , there is a unique ${X \rightarrow Y}$ which factors through ${\ast}$ . Namely, it is

$\displaystyle X \rightarrow \ast \rightarrow Y.$

This is called the zero morphism. In particular, the hom functors take values in the category of pointed sets.

Definition 1 Let ${\mathfrak{C}}$ be a category with zero object. A sequence $\displaystyle A \rightarrow B \rightarrow C$

in the category is called exact if for each ${X \in \mathfrak{C}}$ , the sequence $\displaystyle \hom(X, A) \rightarrow \hom(X, B) \rightarrow \hom(X, C)$

is exact as a sequence of pointed sets. The sequence is said to be coexact if the sequence $\displaystyle \hom(C, Y) \rightarrow \hom(B, Y) \rightarrow \hom(A, Y)$

is exact as a sequence of pointed sets for all objects ${Y \in \mathfrak{C}}$ .

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 ${A \rightarrow B \rightarrow C}$ implies that of ${\hom(X, A) \rightarrow \hom(X, B) \rightarrow \hom(X, C)}$ . So this is something a little different; it’s more analogous to split exactness in standard additive/abelian category theory.

Exact sequences in ${\mathbf{PT}}$

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 ${\mathbf{PT}}$ of pointed topological spaces, it makes sense to talk about exact and coexact things in this category.

First, we have to note that ${\mathbf{PT}}$ has a zero object. Namely, this is the one-point space ${(\ast, \ast)}$ (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 ${f: X \rightarrow Y}$ be a basepoint-preserving morphism of pointed spaces ${(X, x_0), (Y, y_0)}$ . We want to construct a space ${Z}$ with a sequence of pointed spaces

$\displaystyle X \stackrel{f}{\rightarrow } Y \stackrel{i}{\rightarrow} Z$

which is coexact in ${\mathbf{PT}}$ . This means that the composition ${X \rightarrow Z}$ is nullhomotopic and a map out of ${Y}$ extends over ${Z}$ if and only if its pullback to ${X}$ is nullhomotopic.

Here is the construction of ${Z}$ . Namely, we start with the reduced cone of ${X}$ : this is the smash product ${I \wedge X}$ , or alternatively

$\displaystyle CX = X \times I/\left\{ \left\{1\right\} \times X \cup x_0 \times I\right\} .$

This is a pointed space into which ${X}$ embeds; moreover, the inclusion ${X \rightarrow CX}$ 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 ${Z = Y \cup_f CX}$ . This is the space obtained from the disjoint union ${Y + CX}$ by attaching the points ${(x, 0) \in CX}$ to ${f(x) \in Y}$ . Then ${Z}$ is a pointed space, and there is a canonical inclusion ${i: Y \rightarrow Z}$ which preserves basepoints.

Proposition 2 The sequence $\displaystyle X \stackrel{f}{\rightarrow} Y \stackrel{i}{\rightarrow} Z = Y \cup_f CX$

is coexact in ${\mathbf{PT}}$ .

Proof: Indeed, the composite ${i \circ f}$ is nullhomotopic because we can consider the family of embeddings

$\displaystyle g_t: X \rightarrow Y \cup_f CX, \quad g_t(x) = (x,t) \in CX.$

As ${t}$ goes to ${1}$ , this goes into the basepoint of ${Y \cup_f CX}$ .

Conversely, suppose we have a map ${q: Y \rightarrow A}$ nullhomotopy ${Q: X \times I \rightarrow A}$ of ${q \circ f}$ . Then we can use ${q}$ and ${Q}$ together to define a map on ${Y \cup CX}$ into ${A}$ . So if ${q \circ f}$ is nullhomotopic, then it extends to ${Z}$ . This proves coexactness. $\Box$

The Barratt-Puppe sequence

Now we want to extend this coexact sequence further. Well, the natural thing to do would be to write (with ${i}$ denoting the inclusion ${Y \rightarrow Y \cup_f CX}$ ) a sequence:

$\displaystyle Y \cup_f CX \stackrel{j}{\rightarrow} (Y \cup_f CX) \cup_i CY \stackrel{k}{\rightarrow} ((Y \cup_f CX) \cup_i CY) \cup_j (C(Y \cup CX)).$

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 ${(Y \cup_f CX) \cup_i CY}$ . As the figure shows, this is basically the union of two cones such that the end of the cone ${CX }$ is glued into part of the end of ${CY}$ . The “small cone” ${CX}$ is glued into the “big cone” ${CY}$ . Now, the big cone ${CY}$ 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 ${CX}$ , except that its base has been identified to a point. So we get ${CX/\left\{X \times \left\{0\right\}\right\}}$ , which is just the reduced suspension ${\Sigma X}$ discussed earlier.

Proposition 3 The space ${(Y \cup_f CX) \cup_i CY}$ is homotopy equivalent to ${\Sigma X}$ .

This follows from the above quotienting argument, if we grant the assumption that we are allowed to quotient by the contractible guy ${CY}$ . I will come back to why this is allowed later.

We can play the same game for the seemingly more complicated term

$\displaystyle ((Y \cup_f CX) \cup_i CY) \cup_j (C(Y \cup CX))$

and argue that we can quotient by the cone ${C(Y \cup_f CX)}$ . In this case, we are just left with the suspension ${\Sigma Y}$ . Moreover, the map ${\Sigma X \rightarrow \Sigma Y}$ obtained from the map ${k}$ by collapsing these two terms is homotopic to the suspension of ${f}$ , i.e. ${\Sigma f}$ . This takes a little self-convincing, and is best left to the reader.

So we find that our coexact sequence becomes

$\displaystyle Y \cup_f CX \rightarrow \Sigma X \stackrel{\Sigma f}{\rightarrow} \Sigma Y.$

Here the first map is the map that collapses ${Y}$ and ${X \times \left\{-\right\} \subset CX}$ to a point. The second map is the suspension of ${f}$ .

Proposition 4 Let ${f: X \rightarrow Y}$ . Then there is a coexact sequence in ${\mathbf{PT}}$ : $\displaystyle X \rightarrow Y \rightarrow Y \cup_f CX \rightarrow \Sigma X \rightarrow \Sigma Y.$

Recall that a long sequence of elements in ${\mathbf{PT}}$ 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 ${\mathbf{PT}}$ , of the form ${A \rightarrow B \rightarrow B \cup CA}$ in some manner (with ${B, A}$ varying). So we find that this five-term sequence is exact.

Theorem 5 (Barratt-Puppe) Given ${f: X \rightarrow Y}$ , there is a long coexact sequence in ${\mathbf{PT}}$ , going ${X \rightarrow Y \rightarrow Y \cup_f CY \rightarrow \Sigma X \rightarrow \Sigma Y \rightarrow \Sigma (Y \cup_f CX) \rightarrow \Sigma^2 X \rightarrow \dots}$ .

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: $\Box$

Lemma 6 Suppose ${A \rightarrow B \rightarrow C}$ is coexact in ${\mathbf{PT}}$ . Then the suspension ${\Sigma A \rightarrow \Sigma B \rightarrow \Sigma C}$ is coexact as well.

Proof: This follows purely formally because ${\Sigma}$ has a right adjoint ${\Omega}$ . Indeed, let ${X}$ be a space; we show that

$\displaystyle \hom_{\mathbf{PT}}( \Sigma C, X) \rightarrow \hom_{\mathbf{PT}}( \Sigma B, X) \rightarrow \hom_{\mathbf{PT}}( \Sigma A, X)$

is exact as a sequence of sets. However, this sequence is just

$\displaystyle \hom_{\mathbf{PT}}( C, \Omega X) \rightarrow \hom_{\mathbf{PT}}( B,\Omega X) \rightarrow \hom_{\mathbf{PT}}( A, \Omega X) ,$

which is exact by coexactness of ${A \rightarrow B \rightarrow C}$ . This proves the lemma. $\Box$

So we have shown the theorem. The associated sequence is called the Barratt-Puppe or simply Puppe sequence associated to the morphism ${f: X \rightarrow Y}$ .

The loose end

Technically, it remains to explain why we had the homotopy equivalence

$\displaystyle (Y \cup_f CX) \cup CY \simeq (Y \cup_f CX)/CY = \Sigma X.$

In other words, we should explain why we were allowed to quotient by the contractible space ${CY}$ 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 ${(Y \cup_f CX) \cup CY}$ into itself carrying ${CY}$ into a point. Formally, let us define this idea.

Definition 7 Let ${(X, x_0)}$ be a pointed space and ${A}$ a closed subspace containing the basepoint. A deformation of ${X}$ is a homotopy of the identity. That is, it is a continuous map ${H: X \times I \rightarrow X}$ such that ${H(0, \cdot) = 1_X}$ and ${H(x_0 \times I) = x_0}$ . We say that ${H}$ carries ${A}$ into a point if ${H(A \times 1) = \ast}$ and ${H(A \times I ) \subset A}$ .

Now, let us prove:

Proposition 8 Suppose ${(X, x_0)}$ is a pointed space and ${A}$ a closed subspace containing ${x_0}$ . Suppose there is a deformation ${H: X \times I \rightarrow X}$ carrying ${A}$ into ${x_0}$ . Then the quotient map $\displaystyle f: X \rightarrow X/A$

is a homotopy equivalence (homotopies preserving the basepoint).

Proof: Indeed, we define the map

$\displaystyle g: X/A \rightarrow X$

via ${\overline{x} \rightarrow H(\overline{x}, 1)}$ . Note that ${H(\cdot, 1)}$ collapses ${A}$ to a point so factors through ${X/A}$ . I claim that ${f,g}$ are homotopy inverses.

First, consider ${g \circ f: X \rightarrow X}$ . This is the map ${x \rightarrow H(x, 1)}$ . It is homotopic to ${x \rightarrow H(x,0) = x}$ . So this is homotopic to the identity.

Next, consider ${f \circ g: X/A \rightarrow X/A}$ . This is the map ${\overline{x} \rightarrow {H(\overline{x}, 1)}}$ . But ${H}$ factors through ${A}$ because ${H(A \times I) \subset A}$ , so it becomes a map

$\displaystyle \overline{H}: X/A \times I \rightarrow X/A.$

In particular, this map ${\overline{H}}$ is a homotopy between ${f \circ g}$ and ${1_{X/A}}$ . $\Box$

So to finish the details in the proof of the Barratt-Puppe sequence, we just need to check that there is a deformation of ${(Y \cup_f CX) \cup CY}$ that takes ${CY}$ into a point. The idea is to push ${CY}$ down into its “nose” while stretching ${CX}$ slightly. I leave the details to the reader.

Climbing Mount Bourbaki