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We know that there is a map from $h:\pi_{i}^{st}(pt)\rightarrow KO_{i}(pt)$ and we know all the $KO_{i}(pt)$ by Bott periodicy: they are $Z, Z_{2},Z_{2},0,Z,0,0,0$. We also know $\pi_{i}^{st}(pt)$ for i=0~7.

My question is how to determine the image of $h$? In particular (which is the case I care most), how to prove that when $i=1$, the Hurewice image of the hopf map is non zero in $KO_{1}(pt)$?

I can't find this computation in any reference although this is standard. Can anyone tell me about the reference?

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For $i = 1$ you can argue using the fact that $S_n \to O(n)$ induces an isomorphism on abelianizations. $\pi_i^{st}$ are the homotopy groups of the group completion of $\coprod_n BS_n$, and at least for $i$ positive $KO_i$ are the homotopy groups of the group completion of $\coprod_n BO_n$.

For $i = 2$, you can argue that the map is surjective using Pontrjagin-Thom construction and the Atiyah invariant. $\pi_2^{st}$ is the set of cobordism classes of framed surfaces, and the map to $KO_2 = \mathbf{Z}/2$ factors through the "Atiyah invariant'' of spin surfaces, I'm not sure whether that boils down to something elementary.

For $i = 4,8,12,16,20,\dots$ the map is zero because the domain of the map is torsion.

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    $\begingroup$ These 2-torsion classes can be realized by maps from a mod-2 Moore spectrum such as $S^1 \to \Sigma M(2) \to S$, and the Moore spectrum has a map $\Sigma^8 M(2) \to M(2)$ which, upon precomposition, induces the 8-fold periodicity $KO_i(pt) \to KO_{i+8}(pt)$, so your proof actually implies surjectivity onto all the torsion classes of $KO$. $\endgroup$
    – Tyler Lawson
    Commented Jan 2, 2014 at 20:07

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