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Suppose we have two coverings $p_1:Y\rightarrow X$ and $p_2:Y^\prime\rightarrow X$ and further a continuous map $\pi\colon Y\rightarrow Y^\prime,$ such that $p_2\circ\pi=p_1.$

Ist $\pi$ always a covering? If not, what are kind of the minimal prerequisites I have to admit to $\pi$ if I want to?

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    Okay, you're right. I added the property of beeing continuous to $\pi$.2012-12-30

3 Answers 3

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It is a covering if $\pi$ is continuous and onto. Take $y' \in Y'$, we want to find a neighborhood $U_{y'}$ of $y'$ such that the preimage under $\pi$ is the union of some disjoint open sets homeomorphic to $U_{y'}$ under $\pi$.

Let $p_2(y') = x$. There is a neighborhood $V_x$ of $x$, such that 1. $p_2^{-1}(V_x)$ is a disjoint union of open sets, each homeomorphic to $V$ under $p_2$. Call the open set containing $y'$ $U_{y'}$, and $q_2$ to be the inverse (homeomorphism) of $p_2 |_{U_{y'}} : U_{y'} \to V_x$. 2. $p_1^{-1}(V_x)$ is a disjoint union of open sets $W_a$, each homeomorphic to $V$ under $p_1$, and $p_1(a) = x$.

We need $\pi$ to be onto to make sure the preimage $\pi^{-1} (U_{y'})$ isn't empty. If $\pi$ is continuous, then $\pi^{-1}(U_{y'})$ can be checked to be a disjoint union of open sets $W_a$, with $\pi (a) = y'$. Continuity is used to make sure that $W_a$ are mapped to $U_{y'}$, for otherwise it may be possible to split $W_a$ into pieces, each mapping to different covering neighborhood, as shown in the other answer.

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    @beginner, not exactly. To be a counterexample it should have all the hypothesis (on your link from Wiki) but not be a covering. Your map $\pi$ is not onto so it is obvious that it is not a covering.2012-12-30
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It seems you have edited the question to now assume continuity, but let me point out that preserving fibers is in general rather far away from continuity. As a concrete example, let $Y = Y \prime = \mathbb{R}$ and $X = S^{1}$. The covering $p: \mathbb{R} \rightarrow S^{1}$ is the usual one, ie. given by $p(x) = e^{2 \pi i x}$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = x + [x]$, where $[x]$ is the floor function. This is fiber-preserving, since the fibers of $p$ are all of the form $x + \mathbb{Z}$ and $[x]$ and the floor function only takes integer values. But $f$ is not continous, because if it were, then $[x] = f(x) - x$ would also be, but it has "jumps" at all the integers.

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    Yeah, I already got this. Thanks!2012-12-30
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I think we need (at most) locally connectedness (and trivially, surjectivity).

Adopting Sanchez's notations, I think continuity is not sufficient to deduce $W_a$ is mapped to $U_{y'}$. Actually, continuity can't prevent $W_a$ being mapped into distinct $U$s(although one of them must be $U_{y'}$. But if we take $V_x$(and then $W_a$s) to be connected, a single $W_a$ will be mapped into $U_{y'}$ as its image must be connected while those $U$s are disjoint and all open. However, I don't know any counterexample if we don't assume locally connectedness.

Another remark is, under the assumption of locally connectedness, we can require $Y'$ to be connected to guarantee surjectivity. This is because, either, say, the whole of $U_{y'}$ has a preimage, or they all have no. Thus they constitute an open partition of $Y'$, and it follows from connectedness that all of $Y'$ belongs to the former.