$hf\alpha:I\to S^1$ is not null-homotopic I was studying this proof, I understood everything, except the part where $hf\alpha:I\to S^1$ is not null-homotopic.
We have, that $f:S^2 \to S^1$ a continous map such that $\space f(-x)=-f(x)$ $ \space \forall \space x \in S^2$,  $h:S^1 \to S^1$ is the continous map $h(z)=z^2$ and $\alpha : I \to S^2 $ is a path with $\alpha(0)=x$ and $\alpha(1)=-x$, ie, has as extremal points two antipodal points.
In fact I have no idea how to show that a map cannot be null-homotopic, the only one I know is the identity map from $S^1$ to $S^1$.

You can take any path $\alpha$ in $S^2$ from any $x \in S^2$ to its antipodal point $-x \in S^2$. Then $f\alpha$ is a path in $S^1$ from $y = f(x)$ to $-y = -f(x) = f(-x)$. Let $\gamma$ be the path in $S^1$ running with constant speed counterclockwise from $-y$ to $y$. Then $f\alpha * \gamma $ is a closed path which has a winding number $n \in \mathbb Z$. Hence $h(f\alpha * \gamma) = hf\alpha * h\gamma$ is a closed path which has a winding number $2n$. Note that both parts $hf\alpha$ and $h\gamma$ are closed paths, where $h\gamma$ has degree $1$. But $hf\alpha$ is path homotopic to $h(f\alpha * \gamma) * (h\gamma)^{-1}$ and thus has winding number $2n-1$. Hence it cannot be inessential.
