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Let $X,Y$ be complex algebraic surfaces. Is there a map $f : X \to Y$ which is not algebraic but induces an homeomorphism with respect to the Zariski topology ?

Motivation : there are plenty of such maps for curves, since any bijection is automatically an homeomorphism. For surfaces, I suspect that homeomorphism should be given by an algebraic map but I'm not able to prove it.

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Let $X\subset \Bbb A^n_{\Bbb C}$ be a surface defined by $V(f_1,\cdots,f_m)$ where all $f_i\in \Bbb R[x_1,\cdots,x_n]$. Then complex conjugation is a homeomorphism of $X$ with itself in the Zariski topology, but not an algebraic map.

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