Degree of a map $f: S^3 \times S^1 \to S^4$ which is odd with respect to $S^3$

Question

Let $f: S^3 \times S^1 \to S^4$ be a map which satisfies two assumptions:

1. $f$ is antipodal with respect to $x \in S^3$, that is, $$ f(-x;y) = -f(x;y). $$
2. If we use the Cartesian coordinates $x = (x_1,x_2,x_3,x_4) \in S^3$, then $$ f(x_1,x_2,x_3,x_4;-y) = f(x_2,x_1,x_3,x_4;y). $$ That is, first two coordinates $x_1,x_2$ permute.

Question: What is the value of deg$(f)$?

Unfortunately, I have no idea how to attack this question. Any help will be appreciated.

Answer 1

Let \begin{align*} g \colon S^3 \times S^1 &\to S^3 \times S^1\\ (x_1,x_2,x_3,x_4,y) &\mapsto (-x_1,-x_2,-x_3,-x_4,y) \end{align*} i.e. $g=-id_{S^3} \times id_{S^1}$. Then $deg(g)=1$ (because it is a composition of reflections of the first $4$ coordinates, each of degree $-1$). Now $$f(-x,y)=f( g(x,y))=-f(x,y)$$ by the first hypothesis. But then $$deg(f \circ g)=deg(f)deg(g)=deg(f)=-deg(f)$$ so $deg(f)=0$.