The universal covering space of a plane excluding the origin $\mathbb{R}^2 \backslash \{\mathbf{0}\}$ can be regarded as the Riemann surface of the complex-valued function $\ln(z)$. My question is what is the universal covering space of a plane excluding two points $\mathbb{R}^2\backslash\{\mathbf{a},\mathbf{b}\}$? Can it also be written as a Riemann surface of a complex-valued function?

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    $\begingroup$ I think it will be quite crazy.. something like the universal cover of figure $8$ with a line attached to every point.. math.stackexchange.com/questions/354056/… $\endgroup$ Mar 24, 2017 at 11:09
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    $\begingroup$ It may be $ln(z-a)(z-b)$. $\endgroup$
    – Semsem
    Mar 24, 2017 at 11:14
  • $\begingroup$ @SamehShenawy Consider two loops, which respectively wind one of the two holes one time, and share a common starting and end point. Their lifted paths (stating at the same point) on the Riemann surface of $\ln(z−a)(z−b)$ will also have the same end. However, these two lifted paths shouldn't be homotopic, this indicates the Riemann surface is not simply connected, and thus it can't be the universal covering. $\endgroup$
    – zrysky
    Mar 24, 2017 at 13:25


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