I'm curious about the homotopy groups of $S^1 \lor S^1$ and more generally $S^n \lor S^n$. What is the current knowledge about them?
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asked Nov 14, 2022 at 21:12
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3$\begingroup$ The first homotopy group of $S^1\vee S^1$ is easy by a certain theorem involving path connected intersections (don't recall the exact statement) - it should be isomorphic to the free group on two letters, given $\pi_1(S^1)\cong\Bbb Z$ $\endgroup$– FShrikeCommented Nov 14, 2022 at 21:29
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1 Answer
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$S^1 \vee S^1$ has all higher homotopy groups trivial: its universal cover is a tree, which is contractible.
For $S^n \vee S^n$ you are not so lucky. You can still algorithmically compute these homotopy groups using Hilton's theorem if you happen to know all relevant homotopy groups of spheres (nobody knows all homotopy groups of all spheres and algorithmic computation is not very good).
