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Tom Goodwillie
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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.)

Tom Goodwillie
  • 55.9k
  • 7
  • 151
  • 240