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While I was able to prove that there is no surjective, smooth map $f:S^1 \to S^1 \times S^1 \times S^1$ but I am unable to prove this.

Suppose such a map exists, then by sard's theorem, it has a regular value. Hence there exists a unique $x \in S^2$ where $d_xf$ is surjective. But I am not able to proceed further. Any hint?

permutation_matrix
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    Such a map would violate Borsuk-Ulam after composing with the inclusion from the 1-sphere to the 2-sphere. – Randall Jun 09 '21 at 11:51
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    "There exists a unique $x \in S^{2} $ where $d_{x} f$ is surjective" is plain wrong. I think you misunderstand Sard. Anyway, i'd like to offer a possibly simpler solution than Randall's: f is a closed map since it's continuous between compact spaces, so if it's one to one it's an embedding of $S^{2}$ in $S^{1}$. If f is not onto this is a contradiction to the non-contractibility of $S^{2}$, and if f is onto it's a homeomorphism which is a contradiction to, say, $S^{2}$ being simply connected. – Tom Ariel Jun 09 '21 at 13:40

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