I want to show that an continuous even map, i.e., it satisfies $f(z)=f(-z)$, on the circle $\mathbb S^1$ has even degree without any homology techniques.
I know the multiplicative property of degrees, i.e., $\deg(f \circ g ) =\deg(f) \deg(g)$. It is trivial that if $f$ is an even map and $p_2$ is the multiplication map $z\mapsto z^2$, then $f\circ A=f$, $p_2\circ A= p_2$ where $A$ is the antipodal map. I also have a continuous map $g$ which satisfies $\deg(g)=\deg(f)$ and $p_2\circ f = g \circ p_2$. Given these information, how can I proceed?
Any help is appreciated.
Best Answer
Define $g:S^1\rightarrow S^1$ to be the function $$g(z)=f(\sqrt z).$$ The square root is defined up sign, but the fact that $f$ is even guarantees that $g$ is well-defined. In fact $g$ is continuous. For the covering projection $p=\exp(2\pi it(-)):\mathbb{R}\rightarrow S^1$ is a quotient map, and $g\circ p$ is the continuous map $$\mathbb{R}\ni t\mapsto f(\exp(\pi it))\in S^1.$$ Now if $p_2:S^1\rightarrow S^1$ is the map $p_2(z)=z^2$, then $g\circ p_2=f$. Thus $$\deg(f)=\deg(g\circ p_2)=\deg(g)\cdot \deg(p_2)=2\cdot \deg(g).$$