1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $H^2(\mathbb{Z},\mathbb{Z})$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.