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Suppose $\pi : \widetilde X \to X$ is a finite, connected covering, and suppose that there exists a continuous map $f: \widetilde X \to \mathbf R^2$ which is injective on each fibre of $\pi$. Is $\pi$ necessarily a homeomorphism?

In my answer to this question, I proved that it is so if $\mathbf R^2$ is replaced by $\mathbf R$. The reason is that we can continuously single out an element of any finite subset of $\mathbf R$ (for example, the maximum, or the minimum). It appears that the order of $\mathbf R$ plays a significant role in the solution to this question, but that something a little weaker than it is actually involved. Since there is no order on $\mathbf R^2$, I think that the answer to the question with $\mathbf R^2$ is no, but I haven't been able to come up with $\pi, f$ which do not possess this property.

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No. A counterexample is given by taking both of $\widetilde X$ and $X$ to be the unit circle in the complex plane, $\pi$ to be $z\mapsto z^2$, and $f$ to be the identity.

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  • $\begingroup$ +1, this is a simpler example than mine. Though in my example, $\widetilde{X}$ and $X$ aren't homeomorphic at all :) $\endgroup$ – Zev Chonoles Mar 1 '13 at 4:18
  • $\begingroup$ Great! thank you David. $\endgroup$ – Bruno Joyal Mar 1 '13 at 4:20
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No; for example, we can take the covering $\widetilde{X}$

              enter image description here

of $X=\mathbb{S}^1\vee\mathbb{S}^1$

                          enter image description here

and $\widetilde{X}$ clearly has an embedding into $\mathbb{R}^2$ which is injective all over, and hence injective on each fiber, but it is not homeomorphic to $X$.

(pictures from Hatcher's Algebraic Topology)

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