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Under the condition that the spaces (or maybe just the total) are connected and locally path connected, is then the a covering the same as a homeomorphism?

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Are you sure this is what you want to ask? As written, the answer is certainly not: consider for instance the covering map from $\mathbb{R} \rightarrow S^1$. In fact one almost always restricts to covering maps in which both the base and total spaces are connected and locally path connected. –  Pete L. Clark Sep 15 '10 at 11:18
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Actually, what I wanted to ask is why a map of coverings f : E -> E' (p: E -> B and p' : E' -> B being coverings with p'f = f) is a map of coverings (mostly why it is surjective) under the condition that we are dealing with connected and locally path connected spaces. I thought that it might follow from what I asked above but you have proved me wrong. –  Down Sep 15 '10 at 14:04
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Dear Down, To prove the surjectivity, you will have to use connectivity of $E'$ (otherwise the result is not true). You want to show that $f(E) = E'$. Here are some hints: (i) think about what properties you need to prove for $f(E)$ to get this equality. (This is where connectedness will be used). (ii) Prove them using the covering space properties. –  Matt E Sep 15 '10 at 18:49
    
Incidentally, this statement (that $E \to B$ is surjective) is also true if we are working with a fibration. This is actually a hint: use path lifting. –  Akhil Mathew Sep 16 '10 at 2:41
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Dear Down, To prove the surjectivity, you will have to use connectivity of $E'$ (otherwise the result is not true). You want to show that $f(E)=E'$. Here are some hints: (i) think about what properties you need to prove for $f(E)$ to get this equality. (This is where connectedness will be used). (ii) Prove them using the covering space properties.

– Matt E Sep 15 '10 at 18:49

Incidentally, this statement (that $E\to B$ is surjective) is also true if we are working with a fibration. This is actually a hint: use path lifting.

– Akhil Mathew Sep 16 '10 at 2:41

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