Map $f:S^2 \to S^1$ with $f(-x) = -f(x)$

Consider a continuous function $f:S^2 \to S^1$ with $f(-x) = -f(x)$ for all $x \in S^2$. I intend to show that such map doesn't exist using topological covering/lifting theory.

Here my attempts: If we assump that such function $f$ exists then since $S^2$ simply connected and $\mathbb{R}$ is a covering of $S^1$ there exist a lift map $g: S^2 \to \mathbb{R}$ with property $f = p \circ g$. Here $p: \mathbb{R} \to S^1$ is the canonical covering map.

Then I tried to consider a path $\gamma:[0,1] \to S^2$ with property $\gamma(0)= -\gamma(1)$, therefore $\gamma(0), \gamma(1)$ are antipodal.

And therefore (by property of $f$) also holds $f(\gamma(0)) = f(-\gamma(1)) = -f(\gamma(1))$.

I guess that considering this property one can deduce a contradiction by considering a lift $\widetilde{\omega}:[0,1] \to \mathbb{R}$ of the path $\omega:= f \circ \gamma$.

Can anybody explain what here goes wrong/ how to get the desired contradiction? Intiutively I guess that something goest wrong with the uniquess of the lift (see homotopy lifting prop)... but I don't find this last step.

Another attempt of mine would be to cosider the new path $\omega^2$ which is obviously a loop. Can I get a contradiction by considering the lift of it? Should the lift be also a Loop? Why?

Remark: As in a comment stated below this problem is a special case of Borsum Ulam and can be proved rigorously using homological methods. The intention of this question how to prove the statement for $n=2$ with toolbox from elementary lifting properties / covering theory.

Let us assume that there exists a map $f$ as in your question.

We regard $S^1$ as the equator of $S^2$. Then $g = f \mid_{S^1} : S^1 \to S^1$ is inessential because it extends to $D^2$ (identified with the upper hemisphere of $S^2$). We have $g(-z) = -g(z)$. Let $p : \mathbb{R} \to S^1, p(t) = e^{it}$, denote the standard covering map.

Since $g$ is inessential, it lifts to $\tilde{g} : S^1 \to \mathbb{R}$. Define $h(z) = \tilde{g}(z) - \tilde{g}(-z)$. Then $$e^{ih(z)} = e^{i\tilde{g}(z) - i\tilde{g}(-z)} = e^{i\tilde{g}(z)}/e^{i\tilde{g}(-z)} = g(z)/g(-z) = -1 .$$ Hence $h(z) \in \{ (2k+1)\pi \mid k \in \mathbb{Z} \}$. Because $h$ is continuous and $S^1$ is connected, we see that $h(z)$ is constant with value $(2k+1)\pi$ for some $k \in \mathbb{Z}$. Being constant, $h$ has the property $h(z) = h(-z)$ for all $z$. From the definiton we see that $h(-z) = -h(z)$ which implies $h(z) = 0$ for all $z$. Hence $0 = (2k+1)\pi$, i.e. $k = -1/2$, which is absurd.