I'll call the pair of the space and its subspace (A, B) contractible if there is a homotopy $\Phi^t: B \to A$ such that $\Phi^0$ is $\text{Id}_B$ and $\text{Im}$ $\Phi^1$ is a point. For example, the pair $(D^i,$ $\partial D^i)$ of the closed ball and its boundary is contractible.
The first part of my problem is quite simple: if the pair (A, B) is contractible, it's easy to show that in its long exact homotopy sequence $\pi_i(A) \to \pi_i(A, B)$ is monomorphism and $\pi_i(A, B) \to \pi_{i-1}(B)$ is epimorphism.
But after that, I've to prove that $\pi_i(A, B) \simeq \pi_i(A) \oplus \pi_{i-1}(B)$. The key moment is that author uses $\oplus$ symbol here.
The case $i \geq 3$ isn't obvious for me. And it's my first question: why our sequence is splitting? Should we construst a splitting morphism $\pi_{i -1}(B) \to \pi_i(A, B)$?
If $i = 1$ problem is incorrect: $\pi_1(A, B)$ group isn't a group.
My second question is what can I do if $i = 2$? We don't know anything about commutativeness of $\pi_1(B)$ and $\pi_2(A, B)$. And it's easy to show an example where $
\pi_1(B)$ isn't commutative (and $\pi_2(A, B)$ also isn't commutative because there is an epimorphism from $\pi_2(A, B)$ to $\pi_1(B)$). In this case, the symbol $\oplus$ is incorrect. So are both of cases $i = 1$ and $ i=2$ incorrect? Or in the second of them, I should change $\oplus$ to $\times$ and prove that $\pi_i(A, B) \simeq \pi_i(A) \times \pi_{i-1}(B)$? How to do that? I haven't any ideas…
Best Answer
Let's look at our exact homotopy sequence. All morphisms $\pi_n(B) \to \pi_n(A)$ are zeros (because the pair is contractible). It follows from this fact that we have a short exact sequence $0 \to \pi_i(A) \to \pi_i(A, B) \to \pi_{i-1}(B) \to 0$. Now there are two cases.
$i \geq 3$. All groups here are abelian so we can use the splitting lemma. Our goal is to construct a splitting morphism $\Omega: \pi_{i−1}(B) \to \pi_i(A, B)$. We can do something like this: let $\psi$ denote a map $S^n \to B$. We can think about $\psi$ as about a null-homotopic map $S^n \to A$. It's null-homotopic so there is a map $\Omega(\psi): D^{n+1} \to A$ such that $\Omega(\psi)|_{S^n}\equiv \psi$. Now we can easily see that $\Omega$ is the splitting map. And my proposition about existence of isomorphism $\pi_i(A, B) \simeq \pi_i(A) \oplus \pi_{i-1}(B)$ follows from this fact.
$i = 2$. The morphism $\Omega$ which was described above is a correct splitting map also. So due to splitting lemma for non-abelian groups $\pi_2(A, B)$ is a semi-direct product of $\pi_2(A)$ and $\pi_1(B)$.
P.S. I would like to gratefully thank user @freakish for useful discussion.