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I have been thinking on the problem below for a while and I am not sure if it is correct or not. I am trying to see if there exists a counter-example for the problem below.

Problem: Let $f: (S^1)^n \rightarrow S^2$ be a $\mathbb{Z}_2$-equivariant map from $n$-torus to $2$-sphere. Is it true that if $f$ is null-homotopic, then $f$ is $\mathbb{Z}_2$-homotopic to a non-surjective map from $n$-torus to $2$-sphere?

I am now going to consider one special case of the problem. If $f$ is non-surjective, then the problem above is trivial.

Any comments about the problem will be greatly appreciated.

PS. Antipodal $\mathbb{Z}_2$-action are used on both torus and sphere.

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    $\begingroup$ What are the group actions, if I may ask? $\endgroup$
    – David Roberts ♦
    Commented Jun 16 at 4:49
  • $\begingroup$ In the case $n=1$ how do you know that $f$ is not surjective? Does it stem from the null-homotopic property, or do you assume some higher regularity on $f$ than mere continuity? $\endgroup$
    – Loïc Teyssier
    Commented Jun 16 at 11:51
  • $\begingroup$ Antipodal $\mathbb{Z}_2$-action are used on both torus and sphere. $\endgroup$
    – Arash
    Commented Jun 16 at 15:10
  • $\begingroup$ In case n=1, we have a map $f$ from a circle $S^1$ to sphere $S^2$. Since a sphere has more dimensions than a circle, $f$ cannot be a subjective map. $\endgroup$
    – Arash
    Commented Jun 16 at 15:12
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    $\begingroup$ You are right. Then I do not have any proof for the case n=1. $\endgroup$
    – Arash
    Commented Jun 16 at 17:14

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