TITLE: A map homotopic to a $\mathbb{Z}_2$-equivariant map is $\mathbb{Z}_2$-equivariant.
QUESTION [5 upvotes]: Assume an antipodal homeomorphism $\nu_n: S^n \rightarrow S^n$ mapping $x$ to $-x$, (it can be seen as an action of $\mathbb{Z}_2$ on $S^n$).
What I want to prove or reject is:
Suppose $h:S^2 \rightarrow S^1$ is a $\mathbb{Z}_2$-equivariant map. If $g:S^2 \rightarrow S^1$ is homotopic to $h$, then $g$ is a $\mathbb{Z}_2$-equivariant map too.
This claim might be true in general for any $G$-equivariant map on any topological space. The general statement would be: Suppose $X$ and $Y$ are two topological spaces equipped with an action of the group $G$ (i.e. $X$ and $Y$ are $G$-spaces). If $h:X \rightarrow Y$ is a $G$-equivariant map homotopic to the map $g:X \rightarrow Y$, then $g$ is $G$-equivariant too.
I am trying to prove the statement for $G=\mathbb{Z}_2$, $X=S^2$, and $Y=S^1$ but it might be wrong. So, either a proof or a counter example would help me.
I know equivariant homotopy theory is interested in studying the homotopy classes of equivariant maps between $X$ and $Y$ but I guess my question does not fit into this subject because we do not know yet if the map $g$ is $G$-equivariant or not.
I have not seen this question anywhere. This is my own thought. I would appreciate any hint or help.

REPLY [3 votes]: For $X = S^2, Y = S^1$ it is true. The reason is that there do not exist $\mathbb{Z}_2$-equivariant maps in this case. $\mathbb{Z}_2$-equivariant maps $h : S^n \to S^m$ between spheres are called antipodal ($h(-x) = - h(x)$). The Borsuk-Ulam theorem says that the existence of an antipodal map is possible only when $n \le m$.
See also Map $f:S^2 \to S^1$ with $f(-x) = -f(x)$.
In general it is false. As an example take $X = Y  = [-1,1]$ with $\mathbb{Z}_2$-action $t \mapsto -t$. Then $h = id$ is $\mathbb{Z}_2$-equivariant, but the constant map $g(t) = 1$ which is homotopic to $h$ is not.