Let $X_i$ be topological spaces. Why does the map $f:\bigvee_i X_i\rightarrow \prod_i X_i$ induce isomorphisms $\pi^s_*(f)$ on stable homotopy groups?

  • $\begingroup$ Finite number of terms? $\endgroup$ – Randall Oct 5 '18 at 19:37

This is false. Indeed, $X\times Y$ is stably equivalent to $X\vee Y\vee (X\wedge Y)$ and so the map $X\vee Y\to X\times Y$ induces isomorphisms on stable homotopy groups iff the stable homomtopy groups of $X\wedge Y$ are trivial. For a very simple example, let $X=Y=S^0$. Then $X\vee Y$ is discrete with $3$ points so $\pi_0^s(X\vee Y)\cong\mathbb{Z}^2$ while $X\times Y$ is discrete with $4$ points so $\pi_0^s(X\times Y)\cong \mathbb{Z}^3$.

  • $\begingroup$ How do you prove that $X\times Y$ is table equivalent to the wedge you described? $\endgroup$ – user09127 Oct 6 '18 at 20:08
  • 1
    $\begingroup$ You can explicitly construct a homotopy equivalence after suspending once. See for instance section 4.I in Hatcher's Algebraic Topology. $\endgroup$ – Eric Wofsey Oct 6 '18 at 20:11

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