Among the properties of the degree map in Hatcher's Algebraic Topology, there is the property that if $f:S^1 \to S^1$ is not surjective then the degree is $0$.
I'm wondering whether the converse statement is true too – i.e. if $deg f =0$ then $f$ is not surjective.
Intuitively, this seems correct: if we view the degree as the number of times we're wrapping around $S^1$ then if we're not wrapping around even once — we're not surjective.
Trying to think about how to prove this, I reasoned as follows: I would have to show that $f_* :H_1(S^1) \to H_1(S^1)$ is the zero map, which means that I'm mapping (equivalence classes of) cycles $\alpha$ in the original space, to a point (?) in the other space (a point, because $\alpha$ must be in Img$\partial_2$, seeing as it's sent to the $0$ class, but since there are no singular $2$-simplices, then this must be sent to the $0$ element in $C_1(S^1)$, which geometrically might represent a point (?), but algebraically it makes no sense because the elements here are formal sums of singular-$1$ simplices). This is clearly not the way to go about it…
I was unable to come up with a counter example either.
Any ideas how to prove/refute?
Best Answer
There are surjective maps with degree $0$. Let $f : S^1 \to J = [-1,1], f(x,y)= x$, and $g : J \to S^1, g(t) = e^{\pi i t}$. These maps are continuous surjections. Thus $g \circ f$ is a continuous surjection which is inessential because $J$ is contractible.