# Calculate homotopy of $S^2$ relative to great circle using exact sequence

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:

$$\pi_2(A,x_0)\to\pi_2(S^2,x_0)\to\pi_2(X,A,x_0)\to\pi_1(A,x_0)\to\pi_1(X,x_0)$$

which is

$$1\to\mathbb{Z}\to\pi_2(X,A,x_0)\to\mathbb{Z}\to1$$

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.

• Since $\mathbb Z$ is a free group, your sequence splits, and you get $\mathbb Z^2$ as a result. – 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 $$\pi_1(S^1)\rightarrow\pi_2(S^2,S^1)$$.

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