This one will be a quick post. In effect, we continue with last time, where we defined the relative homotopy groups, and now describe a practical means of determining when something in one of these relative groups is zero or not. This will become useful in the future.

The compression criterion

We have defined the group ${\pi_n(X, A)}$ above, but we still need a good criterion for knowing when something in ${\pi_n(X, A)}$, represented by ${f: (D^n, S^{n-1}) \rightarrow (X, A)}$ , is zero. Or, when ${n = 1}$, when it represents the base element. The obvious reason is that if there is a homotopy ${H: (D^n, S^{n-1}) \times I \rightarrow (X, A)}$ starting with ${f}$ and ending at the constant map. Here is another that will be useful.

Theorem 1 (Compression criterion) A map ${f: (D^n, S^{n-1}) \rightarrow (X, A)}$ represents zero in ${\pi_n(X, A)}$ if and only if ${f}$ is homotopic relative ${S^{n-1}}$ to a map ${g: D^n \rightarrow A}$.

Proof: This is one of those things which is not really all that hard to prove, but for which pictures help significantly. So I will try to draw pictures. (more…)

Today, we will define relative versions of the homotopy groups, and show that they fit into an exact sequence. So let ${X}$ be a pointed space and ${A \subset X}$ a subspace containing the basepoint. Let ${n \geq 1}$. Then we define

$\displaystyle \pi_n(X, A)$

to be the pointed homotopy class of maps ${(D^n, S^{n-1}) \rightarrow (X, A)}$. Here ${D^n}$ has a basepoint, which is located on the boundary ${S^{n-1}}$.

Definition 1 ${\pi_n(X, A)}$ is called the ${n}$-th relative homotopy group of the pair ${(X, A)}$. (We have not yet shown that it is a group.)

Another perhaps more geometric way of thinking of the relative homotopy groups is as follows. Namely, it is the homotopy class of maps ${(I^n, I^{n-1}, J^n) \rightarrow (X, A, x_0)}$, where ${I^n}$ is the ${n}$-cube and ${J^n}$ is the complement of the front face ${I^{n-1}}$. So on the boundary, such a map is ${x_0}$ except possibly on the front face, where it is at least in ${A}$. The reason is that if we quotient by ${J^{n-1}}$, we get the pair ${(D^n, S^{n-1})}$. (more…)

The Whitehead theorem states that a map of connected CW complexes that induces an isomorphism in homotopy groups is a homotopy equivalence. In particular, isomorphisms in the homotopy category of pointed CW complexes can be detected by homming out of spheres $S^n$. But the equality of two morphisms cannot. The fact that this “relative Whitehead theorem” fails was the subject of a MO question. Today, I want to discuss another example along these lines. (I will assume a little more familiarity with algebraic topology than I have in previous posts.)

Recall that a common technique to show that a map is not nullhomotopic is to show that it does not induce the trivial morphism on some functor in algebraic topology. For instance, the fact that ${\pi_1(S^1) \neq 0}$ is used to show that ${S^1}$ is not contractible; this is probably the most basic example. But the basic invariants of algebraic topology can be insufficient. Here is an example which Eric Larson showed me yesterday.
(more…)

Last time, we defined two functors ${\Omega}$ and ${\Sigma}$ on the category ${\mathbf{PT}}$ of pointed topological spaces and (base-point preserving) homotopy classes of base-point preserving continuous maps. We showed that they were adjoint, i.e. that there was a natural isomorphism

$\displaystyle \hom_{\mathbf{PT}}(X, \Omega Y) \simeq \hom_{\mathbf{PT}}(\Sigma X, Y).$

We also showed that ${\Omega Y}$ is naturally an H group, i.e. a group object in ${\mathbf{PT}}$, for any ${Y}$. So, given that we have a group operation ${\Omega Y \times \Omega Y \rightarrow \Omega Y}$, it follows that ${\hom_{\mathbf{PT}}(X, \Omega Y)}$ is naturally a group for each ${Y}$. (more…)