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.

Suppose first that ${f}$ is homotopic relative to ${S^{n-1}}$ to something taking values in ${A}$. Then, without loss of generality, we can assume that ${f(D^n) \subset A}$ itself; under this assumption, we must show ${f}$ represents zero in the relative homotopy group. But ${D^n }$ is contractible to its basepoint. So ${f}$ is homotopic to a constant map (relative to the basepoint). But throughout the contraction, the modified ${f}$‘s still stay in ${A}$. So ${f}$ represents the same as the map sending ${D^n}$ to a basepoint, i.e. ${f}$ is trivial.

Now suppose ${f}$ is trivial. Then there is a homotopy as in the figure, a map from the cylinder into ${X}$ which is in ${A}$ on ${S^{n-1} \times I \cup D^n \times 1}$.

This is a homotopy between ${f}$ and the constant map which at each stage sends the boundary ${S^{n-1}}$ into ${A}$. Now we can deform ${f}$ in this homotopy by shifting the top disk down but keeping its “arms” at the boundary.

The end result is that ${f}$‘s boundary changed but the final deformation has values in ${A}$ only. This proves the converse. $\Box$

In fact, the compression criterion alone can be used to prove the exactness of the homotopy group sequence.

${n}$-connected pairs

With the compression criterion in mind, we now formulate the notion of an ${n}$-connected pair.

Definition 2 The pair ${(X, A)}$ (with basepoint ${x_0}$) is said to be ${n}$-connected if$\displaystyle \pi_i(X, A) = 0, \quad i \leq n.$

We can draw an exact sequence of homotopy groups at each stage:

$\displaystyle \pi_{i+1}(X, A) \rightarrow \pi_i(A) \rightarrow \pi_i(X) \rightarrow \pi_i(X, A) \rightarrow \dots.$

As a result, it follows that if ${(X, A)}$ is ${n}$-connected, then

$\displaystyle \pi_i(A) \rightarrow \pi_i(X)$

is an isomorphism for ${i< n}$ and surjective for ${i=n}$. Conversely, if this condition holds, then the pair is ${n}$-connected.

The compression criterion now gives us a useful reformulation:

Proposition 3 The pair ${(X, A)}$ is ${n}$-connected if every ${(D^n, S^{n-1}) \rightarrow (X, A)}$ is homotopic relative the boundary to a map ${D^n \rightarrow A}$.

We will give examples of this in the future.