Let $(X, d_1),(Y, d_2)$ be metric space, $f:X\rightarrow Y$ is called covering map, if for evry $y\in Y$, there is open set $U$ of $y$ such that $f^{-1}(U)$ is a union of disjoint open sets in $X$, such that each of which is mapped homeomophically on to $U$ by $f$
Question 1: Suppose $(X, d_1)$ is compact metric space and $f:X\rightarrow Y$ is surjective continuous map.
$f:X\rightarrow Y$ is a covering map?