Let $A\cong S^1$ be a great circle of $S^2$. I would like to calculate the relative homotopy $\pi_2(S^2,A,x_0)$. I know I have a long exact sequence of relative homotopies of pairs:


which is


As per Hatcher, the map $\mathbb{Z}\to\pi_2(X,A,x_0)$ should be induced by the inclusion $A\hookrightarrow X$, but I don't know what that implies.

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    $\begingroup$ Since $\mathbb Z$ is a free group, your sequence splits, and you get $\mathbb Z^2$ as a result. $\endgroup$ – Pedro Tamaroff Nov 9 '18 at 12:36

Recall that for a pair of spaces $(X,A)$ with fixed baspoint $\ast\in A$, the set $\pi_n(X,A,\ast)$ can be interpreted as the set of relative homotopy classes of maps $(D^n,S^{n-1},\ast)\rightarrow (X,A,\ast)$. For expositional ease I'll suppress the basepoints from further notaion. Now for $n\geq 2$, the set $\pi_n(X,A,\ast)$ has a group structure, and in the long exact sequence of the pair, at the point

$$\dots\rightarrow\pi_2(X)\rightarrow \pi_2(X,A)\rightarrow \pi_1(A)\rightarrow\dots,$$

the image $\pi_2X$ in $\pi_2(X,A)$ is a central subgroup.

In the case at hand we have $X=S^2$ and $A=S^1$, which gives us a short exact sequence of groups

$$0\rightarrow\pi_2(S^2)\rightarrow\pi_2(S^2,S^1)\rightarrow \pi_1(S^1)\rightarrow1.$$

As noted above, this is a central extension of $\pi_1(S^1)\cong\mathbb{Z}$ by $\pi_2(S^2)\cong\mathbb{Z}$, and such gadgets are classified by the elements in the group cohomology $H^2(\mathbb{Z},\mathbb{Z})$. However, it is well-known that $H^2(\mathbb{Z},\mathbb{Z})=0$,so the only such extension is the trivial extension. Hence the sequence is split and

$$\pi_2(S^2,S^1)\cong \mathbb{Z}\oplus\mathbb{Z}.$$

Another way to see this result uses the fact that the inclusion $i:S^1\hookrightarrow S^2$ is null-homotopic, since $\pi_1S^2=0$. Hence the homotopy fibre $F_i$ of $i$ has the trivial homotopy type $F_i\simeq S^1\times\Omega S^2$ and there is a homotopy fibration seqnce

$$\Omega S^2\hookrightarrow S^1\times \Omega S^2\xrightarrow{pr} S^1\rightarrow S^2.$$

Recalling that in general, for $i:A\hookrightarrow X$ the inclusion, there is an isomorphism $\pi_n(X,A)\cong \pi_{n-1}F_i$, which leads us to the same result

$$\pi_2(S^2,S^1)\cong\pi_1(S^1\times\Omega S^2)\cong \pi_1(S^1)\oplus\pi_1(\Omega S^2)\cong\pi_1(S^1)\oplus\pi_2(S^2)\cong\mathbb{Z}\oplus\mathbb{Z}.$$

If you want to be more explicit about it, you can use the fact that for a map $\alpha:S^1\rightarrow S^1$, a null-homotopy of $i_*\alpha\in\pi_2S^2$ is equivalent to an extension $\tilde\alpha:D^2\rightarrow S^2$. Then using a bit of point-set topology to make sure your the extension $\tilde\alpha$ has the right boundary conditions, it is exactly the element in $\pi_2(S^2,S^1)$ under the first description of this group that maps to $\alpha$ under the boundary homomorphism $\partial:\pi_2(S^2,S^1)\rightarrow\pi_1(S^1)$. From here it is not difficult to verify explcitly that the assignment $\alpha\mapsto \tilde\alpha$ can be made to respect the group products, and so is the splitting homomorphism $\pi_1(S^1)\rightarrow\pi_2(S^2,S^1)$.

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  • $\begingroup$ $H^2(\mathbb{Z},\mathbb{Z})$ only classifies central extensions of $\mathbb{Z}$ by $\mathbb{Z}$. In general relative $\pi_2$ is not abelian. $\endgroup$ – Najib Idrissi Nov 9 '18 at 14:24
  • $\begingroup$ @NajibIdrissi, the image of $\pi_2(S^2)$ in $\pi_2(S^2,S^1)$ is a central subgroup, as I pointed out. $\endgroup$ – Tyrone Nov 9 '18 at 14:50
  • $\begingroup$ ... you're right, of course. Sorry. $\endgroup$ – Najib Idrissi Nov 9 '18 at 15:02

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