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.



The second stable homotopy group
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.


