In Hatcher, a covering space of a space $X$ is a space $\tilde{X}$ together with a map $p:\tilde{X} \rightarrow X$ such that for each $x \in X$ there is an open neighborhood $U$ of $x$ such that $p^{-1}(U)$ is the disjoint union of open sets in $\tilde{X}$, each of which is mapped homeomorphically onto $U$ by $p$. It says in the same paragraph "We allow $p^{−1}(U)$ to be empty, the union of an empty collection of sheets over $U$, so $p$ need not be surjective."

However, in several other definitions I see (in Wikipedia, for example: https://en.wikipedia.org/wiki/Covering_space#Universal_covers), $p$ is defined as a "continuous surjective map". So given Hatcher's definition, I'm wondering under what situations we can assume that $p$ is surjective. For example, given that $X$ is path-connected, locally path-connected, and semilocally simply connected, do we know that the universal cover $p:\tilde{X} \rightarrow X$ is surjective?

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    $\begingroup$ A covering space/covering map is not the same as the universal cover... $\endgroup$ – David C. Ullrich Mar 16 '16 at 15:22
  • $\begingroup$ @DavidC.Ullrich Okay, right; the universal cover covers any connected cover; that is, it is the composition of two covering maps. So you are saying the universal cover isn't itself a covering space of $X$? Also, under what conditions can we say that the universal cover is surjective? $\endgroup$ – QuantumDots Mar 16 '16 at 15:42
  • $\begingroup$ My point was that the universal cover is always surjective, by definition; the fact that Hatcher allows non-surjective covering maps does not contradict this, since those maps are not the universal cover... $\endgroup$ – David C. Ullrich Mar 16 '16 at 15:46

A general statement is : A nonempty covering space $\tilde{X}$ of a path connected space $X$ is mapped surjectively.

To see this, use the path lifting property. Since $\tilde{X}$ is nonempty, we can take a point $x_0\in X$ in its image. For any point $y$, we can take a path from $x_0$ to $y$, since $X$ is path connected. Then lift this path to $\tilde{X}$, the end point of the lift is mapped to $y$. Hence the map $p$ is surjective.

You can do better by only assuming $X$ to be connected. See Is a path connected covering space of a path connected space always surjective?


You may see an exercise in covering space section of Hatcher. See errata of Hatcher algebraic topology in the following link. He added surjectivity in that problem.



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