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We have the following short exact sequence from the long exact sequence for a pair $$0\to\pi_2(\mathbb{S^2})=\mathbb{Z}\to\pi_2(\mathbb{S}^2,X)\to\pi_1(X)=F_2\to0.$$ I wanted to construct a section (I guess there is one), so $\pi_2(\mathbb{S}^2)=\mathbb{Z}\rtimes F_2$. What I did is that I tried to use the homotopy extension property for $(\mathbb{D}^2,\mathbb{S}^1)$ to say that I can map a homotopy class in $\pi_1(X)$ to maps that are still homotopic in the set of maps $\mathbb{D}^2\to\mathbb{S}^2$ and whose restrictions to $\mathbb{S}^1$ are also homotopic, so I get a section.

But say, the section $s:\pi_1(X)\to\pi_2(\mathbb{S}^2,X)$ maps a class $[f]\in\pi_1(X)$ to $[F]\in\pi_2(\mathbb{S}^2,X)$ where $F$ is an extension of $f$. Later I think not every map in $[F]$ restricted to $\mathbb{S}^1$ is homotopic to $f$... And if the homotopy extension property really works, the boundary map would always have a section, which seems absurd, though I am totally new to the relative homotopy groups... Even if the group is really a semidirect product, I have no idea how to figure out the action of $F_2$ on $\mathbb{Z}$. My guess is something like $\varphi\in\text{Aut}(\mathbb{Z})$, $\varphi(*^{d_1}*^{d_2}\cdots*^{d_k})=\sum d_i$, where $*^{d_1}*^{d_2}\cdots*^{d_k}$ denotes a word in $F_2$.

Hope to get some help... Thanks!

ah--
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1 Answers1

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Since $\pi_2(A)$ is central in $\pi_2(B,A)$, it is never the nontrivial semidirect product. [See Spanier, Algebraic Topology, Chapter 7, Section 3, Corollary 13, page 386]

The fact that your sequence splits follows formally because $F_2$ is a free group.

Since you mentioned the more general question of the sequence splitting, if we have an exact sequence

$$0 \to \pi_2(A) \to \pi_2(A,X) \to \pi_1(X) \to 0$$

I think we can say the sequence is split if $X \to A$ is nulhomotopic ("$X$ is nulhomotopic in $A$"). This way you can choose a nulhomotopy and use it to make a filling-in of each path, and also a homotopy of their homotopies.

I think this agrees with Tyler Lawson's comment here https://mathoverflow.net/questions/294006/conditions-for-the-second-homotopy-group-to-be-abelian

Ben
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  • Aha! That is indeed beyond my progress on reading Spanier... Thanks! But I am still wondering if there is any example such that the boundary map has no section... – ah-- Dec 07 '18 at 22:18
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    @chikurin No problem :) your last question is a little out of my depth in topology but I'll suggest a line of thought. There is a functor from pairs of pointed spaces to crossed modules (maybe its the same as the one taking the central extension above to the corresponding crossed module) - cf. answers posted on MSE by Ronnie Brown. So you can first find a central extension crossed module that is not split, and try to represent it topologically. – Ben Dec 09 '18 at 11:41
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    Also, just to check - the question your asking is: does there exist a pair of spaces $X \to S$ (with $X$ a subspace) such that $\pi_1(X) \to \pi_1(S)$ is the zero map and $\pi_2(S,X) \to \pi_1(X)$ is split? In particular you don't insist $S = S^2$ I think. – Ben Dec 09 '18 at 11:47
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    @chikurin One problem in the above line is that you will have to find out the corresponding condition to $\pi_1(X \to S^2) = 0$ for crossed modules, or modify your problem to ask when $\pi_2(S,X) \to \text{im}(\delta)$ is split. – Ben Dec 09 '18 at 11:49
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    @chikurin Finally, it's worth mentioning that there is a correspondence between 2-types and crossed modules (and strict 2-groupoids and weak 2-groupoids). This might be relevant, but I don't know how compatible it is with the relation to pairs of spaces from above. – Ben Dec 09 '18 at 11:51
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    @chikurin I haven't studied crossed modules in any depth, so if you are stuck, you also might consider asking this as a separate question as it will be more likely to get attention from the good topologists on here. – Ben Dec 09 '18 at 11:54