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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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What is your definition of covering map? Continuous? Onto? What else? – Sigur Dec 30 '12 at 14:22
Let's take just the definiton of – max Dec 30 '12 at 14:24
But you wrote not necessarily continuous?! How $\pi$ could be a covering map? – Sigur Dec 30 '12 at 14:26
Okay, you're right. I added the property of beeing continuous to $\pi$. – max Dec 30 '12 at 14:29
up vote 1 down vote accepted

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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Are you supposing $\pi$ continuous and onto? – Sigur Dec 30 '12 at 14:34
What is if $Y=X=\{x\},Y^\prime=\{x,y\},p_1=id,p_2=\text{const}$ and $\pi$ the inclusion? So $\pi$ is not surjective. Is this example wrong? – max Dec 30 '12 at 14:39
@beginner, Following the definition you gave (from Wiki) the point $y$ does not have the neighbourhood evenly covered by $\pi$. – Sigur Dec 30 '12 at 14:43
@Sigur, ah I see. Thanks! – user27126 Dec 30 '12 at 14:44
@Sigur Yes, so this is a counterexample to $\pi$ being a covering in my situation, isn't it? – max Dec 30 '12 at 14:46

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! – max Dec 30 '12 at 14:42

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.

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