Existence of square root of identity map on the circle

Question

I am self-studying covering spaces. I am stuck on the following problem.

Does there exists a continuous map $f:S^1\to S^1$ such that $f(x)^2 = x$ for all $x\in S^1$?

I am trying to show the non-existence of this kind of map by lifting it to some covering spaces but was not able to succeed.

Any help will be appreciated. Thanks in advance.

Answer 1

If $f(x)^2 = x$, then $(\times 2) \circ f_* = \mbox{Id}_{\mathbb{Z}}$ is the induced action on the fundamental group, but the $\times 2$ map is not injective, so it can't possible compose with something to give the identity on $\mathbb{Z}$.