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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
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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?

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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.

https://www.math.cornell.edu/~hatcher/AT/AT-errata.pdf

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