The second stable homotopy group

Question

I am interested in the low-degree stable homotopy group $\pi_2^{s}(X)$ of a path-connected space $X$. Using the Atiyah-Hirzebruch spectral sequence, we have the short exact sequence $0\to H_1(X,\mathbb{Z}/2)\to \pi_2^{s}(X)\to H_2(X,\mathbb{Z})\to 0$.

Question: Does it always split?

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Edit: I've been working on a solution to my question, and I believe I've made progress. However, I've hit a roadblock in the proof and I'm hoping for some assistance. I will award the best answer to whoever can help me finish my proof or provide insight into my original question.

I think my exact sequence splits canonically! My idea is to show the following dual exact sequence splits canonically: $$ 0\to H^2(X,\mathbb{Q}/\mathbb{Z})\to\mathrm{Hom}_{\mathbb{Z}}(\pi_2^s(X),\mathbb{Q}/\mathbb{Z})\stackrel{h}{\to} H^1(X,\mathbb{Z}/2)\to 0, $$ where I use the canonical isomorphism $\mathrm{Hom}_{\mathbb{Z}}(H_1(X,\mathbb{Z}/2),\mathbb{Q}/\mathbb{Z})\cong H^1(X,\mathbb{Z}/2)$. Next, I define $\phi$ as the generator of $\mathrm{Hom}_{\mathbb{Z}}(\pi^s_2(K(\mathbb{Z}/2,1)),\mathbb{Q}/\mathbb{Z})\simeq\mathbb{Z}/2$. For each $f\in H^1(X,\mathbb{Z}/2)\cong[X,K(\mathbb{Z}/2,1)]$, I consider the pullback $f^*\phi\in\mathrm{Hom}_{\mathbb{Z}}(\pi^s_2(X),\mathbb{Q}/\mathbb{Z})$. This defines a map $[\phi]:H^1(X,\mathbb{Z}/2)\mapsto\mathrm{Hom}_{\mathbb{Z}}(\pi^s_2(X),\mathbb{Q}/\mathbb{Z})$ that sends $f$ to $f^*\phi$.

I'm tempted to believe that $[\phi]$ is a homomorphism and $[\phi]\circ h=\mathrm{id}_{H^1(X,\mathbb{Z}/2)}$, but I cannot come up with a proof. If you have any insights into this, I would greatly appreciate it. Thank you!

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Edit2 I’m really happy to see some of you enjoyed this question! Then what’s left for myself is to figure out how my argument fails, especially the naturality part.

Answer 1

I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.

Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram $$ \begin{array}{cccccccc} 0&\to& H_1(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0 \\ \downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Hence, if the bottom sequence splits, the upper one too.

Let $A=H_1(X,\mathbb{Z}/2)$. We have another commutative diagram where all vertical maps are isomorphisms: $$ \begin{array}{cccccccc} 0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\ \downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\ 0&\to& H_1(Y,\mathbb{Z}/2)&\to& \pi_2^{st}(Y)&\to&H_2(Y,\mathbb{Z})&\to&0 \end{array} $$ Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. This actually holds for any $Y=K(A,1)$ with $A$ abelian. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

The top sequence in the second commutative diagram splits because $A=H_1(X,\mathbb{Z}/2)$, so it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.

We can also proceed without using the Brown-Loday paper, as hinted below by Tom Goodwillie in a comment. It suffices to show that $\pi_2^{st}(Y)$ is a $\mathbb{Z}/2$-vector space. We have $\pi_2^{st}(Y)=\pi_4(\Sigma^2Y)$ since $Y$ is connected. The space $Y$ is a product of copies of $\mathbb{R}P^\infty$. By the splitting of the suspension of a product, $\Sigma^2 Y$ is a wedge of copies of $\Sigma^2(\mathbb{R}P^\infty\wedge\stackrel{n}\cdots\wedge\mathbb{R}P^\infty)$. The latter space is $4$-connected for $n>2$, hence $\pi_2^{st}(Y)$ is a direct sum of copies of $\pi_4\Sigma^2\mathbb{R}P^\infty=\mathbb{Z}/2$ and $\pi_4\Sigma^2(\mathbb{R}P^\infty\wedge\mathbb{R}P^\infty)=\mathbb{Z}/2$.

P.S. A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.