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I think I knew the answer to this once, but it's Friday afternoon here after a long week...

Let $X$ be a connected space. I want to study the stable Hurewicz homomorphism $$ h\colon\thinspace\pi_1^S(X) \to H_1(X;\mathbb{Z}). $$ I can see using the Atiyah-Hirzebruch spectral sequence that this is an epimorphism (I can also see this using the fact that the stabilisation homomorphism $\pi_1(X)\to \pi_1^S(X)$ factors through the abelianisation $\pi_1(X)\to H_1(X)$, which gives a splitting of $h$).

The spectral sequence also tells me that the kernel is either trivial or $\mathbb{Z}/2\mathbb{Z}=\pi_1^S({\rm pt})$.

I have vague memories of showing that the kernel is always $\mathbb{Z}/2\mathbb{Z}$, but I can't remember how. So I am half expecting an answer of "never" to the question

When is $h$ a monomorphism?

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If by $\pi_1^{st}X$ you mean the unreduced version (as you must when you say that $\pi_1^{st}(pt)=\mathbb Z/2$), then for a based space $X$ you can functorially split off $\pi_1^{st}(pt)$ from $\pi_1^{st}X$, so there is always that kernel.

If you mean the reduced version, then you're precisely talking about the other factor in that splitting.

If we define $\pi_k^{st}X$ as the direct limit of $\pi_{k+n}(\Sigma^n X)$, then we get the reduced version. (Your $\mathbb Z/2$ may be considered as the unreduced $\pi_1^{st}$ of a point, or as the reduced $\pi_1^{st}$ of a $0$-sphere.)

As for the spectral sequence: You probably mean the one that goes from $H_i(X;h_j(X))$ to $h_{i+j}(X)$ (everything unreduced) for any generalized homology theory. Here $X$ is not necessarily based or connected. If $X$ is connected then, choosing a basepoint, you can split off the left ($i=0$) column, so to speak. Or work with the analogous spectral sequence that goes from $H_i(X,pt;h_j(X))$ to $h_{i+j}(X,pt)$ and lacks that column.

That applies to any $h$. By looking separately at all components (or by choosing one point in each component in each component and then noting that this discrete set is a retract of $X$) you see that in general the canonical map $H_0(X;\pi_j(pt))\to h_j(X)$ is a naturally split injection.

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  • $\begingroup$ You are right of course, thank you. It must be nearly pub time. $\endgroup$ – Mark Grant Nov 11 '11 at 16:24

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