# Why is the map from a wedge to a product induces isomorphisms between stable homotopy groups

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?

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

## 1 Answer

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 $$\pi_0^s(X\times Y)\cong \mathbb{Z}^3$$.

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