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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?

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No, not necessarily. Think of an interval folded to cover an interval whose length is a half of it.

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