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I have the following question: Let $\mathbb{D}$ denote the unit disk. Let $f:X_1 \longrightarrow X_2$ be a continuous mapping between Riemann Surfaces. Let $ \pi_1 : \mathbb{D} \longrightarrow X_1$ , and $ \pi_2 : \mathbb{D} \longrightarrow X_2$ be the universal covering spaces of $X_1$ and $X_2$, respectively. A lifting of $f$ is a continuous map $ \tilde{f}: \mathbb{D}\longrightarrow \mathbb{D}$ such that $f\circ \pi_1=\pi_2\circ \tilde{f}.$

The question is to show if $f$ is homeomorphism, then so is $\tilde{f},$ or to give a counterexample.

Any help will be appreciated.

Thank you.

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Yes, $\tilde f$ is a homeomorphism. But let me begin with two remarks.
1) The map $\tilde f$ is not well defined, since you can compose any chosen $\tilde f$ with an automorphism of the covering $\pi_2$. The correct context is that of pointed coverings, covering spaces with a distinguished point chosen.

2) More importantly, the result is easier to understand in a general context, so let us forget about disks and Riemann surfaces! [In what follows, all topological spaces are assumed connected and locally pathwise connected.]

Crucial property A pointed covering map $\pi: (\tilde X, \tilde x_0)\to (X, x_0)$ with $\tilde X$ simply connected is universal in the following sense:
Given any pointed covering $\rho :(\hat Y,\hat y_0) \to (Y, y_0)$ and any pointed continuous map $f:(X,x_0) \to (Y,y_0)$ , there exists a unique morphism of pointed coverings $\tilde f:(\tilde X, \tilde x_0)\to (T,t_0)$ [meaning, of course, that $\tilde f$ is continuous, that $\tilde f (\tilde x_0)=\hat y_0$ and that $\rho \circ \tilde f=f\circ \pi$].

If $f$ is a homeomorphism with inverse $g$, the uniqueness property above (=functoriality) will immediately imply that $\tilde f$ is a homeomorphism with inverse $\tilde g$, which answers your question in the particular case you are interested in.

Bibliography You will find a more general version of the Crucial property on page 61 of Hatcher's Algebraic Topology (Proposition 1.33)

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