One degree self map on $S^1$ but is not diffeomorpism

Question

I want to find a map $F:\mathbb{S}^1\to\mathbb{S}^1$ such that $\text{deg}(F)=1$ and $F$ is not a diffeomorphism. Could $F$ be $F(x,y)=(|x|,y)$?

Answer 1

Your example doesn't work because it's not surjective. Any curve you push foward with $F$ will be a curve in a punctured circle, which is the line. So they will have trivial homology. In terms of explicit examples of $F$, my first thought was the use cohomology:

It's sufficient to find a function $F$ so that $F^*d\theta$ has the cohomology generated by $d\theta$ but with some zeros (as a 1-form). Here, I'm thinking that $\theta$ is the identity on the real line modulo integer multiples of $2\pi$.

E.g, setting $\nu = f d\theta $, the equation $F^*d\theta = \nu$ boils down to $F' = f$. One can take $f = \cos(2\theta) + 1$, which has a solution $F =-1/2 \sin{2\theta} + \theta$ on our circle. Note that $f$ has integral $2\pi$ on our circle and has zeros. Thus $F$ has your properties.