Whether covering transformation of universal covering have not fixed point?

Question

Assuming $\pi: \tilde M\rightarrow M$ be a universal covering of a complete Riemannian manifold $M$.

$f:\tilde M \rightarrow \tilde M$ is a covering transformation. If $f$ has fixed point, whether $f$ must be the identity?

Answer 1

As Moishe Kohan comments, the reason lies in a very general property of covering maps. This is the unique lifting property:

Given a covering map $p : \tilde X \to X$ and a map $\phi : Y \to X$ living on a connected space $Y$, then any two lifts $\tilde \phi_1, \tilde \phi_2 : Y \to \tilde X$ of $\phi$ which agree at one point of $Y$ agree on all of $Y$.

This theorem can be found in any textbook dealing with covering maps. See for example Proposition 1.34 in Hatcher's "Algebraic Topology".

Now observe that your $f : \tilde M \to \tilde M$ is a lift of $\pi : \tilde M \to M$. Since $\pi$ is a universal covering, $\tilde M$ is (pathwise) connected.