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Is the statement above true for $X \cong_{bihol} Y$? I would say yes, since I can transform any holomorphic function on a open set in $X$ to one in $Y$ and vice versa.

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  • $\begingroup$ Your second sentence suggests you meant the geometric genus defined as the dimension of the vector space of holomorphic 1-forms, so you need to inject those 1-forms from X to Y $\endgroup$ – reuns Feb 23 at 20:51
  • $\begingroup$ Yes, I meant $g:= dim H^1(X, \mathcal{O})$. Why 1-forms? $\endgroup$ – User1 Feb 23 at 21:37
  • $\begingroup$ I'm sorry, I don't quite get it yet, $\mathcal{O}$ is the sheaf of holomorphic functions, why do I need to look at 1-forms at all? $\endgroup$ – User1 Feb 24 at 8:36
  • $\begingroup$ Fix some $x_0\in X$, let $R_X$ be the space of functions analytic around $x_0$ having an analytic continuation over any curve $\in X$ (it is the same as the analytic functions on the universal cover). Let $O_X$ be the globally analytic functions on $X$, and $S_X$ those $\in R_X$ whose continuation over any closed loop differs by some element of $O_X$, and $\Omega_X$ the analytic 1-forms. If $X$ is a compact Riemann surface then $O_X$ is just the constant functions and $f\mapsto df$ is an isomorphism $H^1(X,O)=S_X/O_X\to\Omega_X$. If $X,Y$ are biholomorphic those things stay the same on $Y$ $\endgroup$ – reuns Feb 25 at 3:58
  • $\begingroup$ Now I see that in your first comment you wrote "vector space of holomorphic one forms", I kind of overread that but that's not my definition of genus, its the dimension of the first cohomology group for the golomorphic functions (not 1- forms) $\endgroup$ – User1 Feb 25 at 9:34
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Biholomorphic $\implies$ homeomorphic...

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