Timeline for Is $\mathbb{C}^2$ homeomorphic to $\mathbb{C}^2 - (0,0)$ with the Zariski topology?

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Mar 4, 2017 at 14:41	comment	added	Zach Teitler		@UriBader: $\mathbb{P}^2$, $\mathbb{P}^2-\{P\}$, $\mathbb{A}^2$ are pairwise nonhomeomorphic because, as argued in the penultimate paragraph of [2], the latter two spaces contain disconnected (field other than $\mathbb{C}$!) irreducible curves while $\mathbb{P}^2$ does not; and $\mathbb{P}^2-\{P\}$ contains curves that meet every other curve (e.g., any curve not through $P$) while every curve in $\mathbb{A}^2$, say $f=0$, is disjoint from some other curve, such as $f=1$. I don't see how it helps for $\mathbb{A}^2-\{P\}$.
Mar 4, 2017 at 8:05	comment	added	Uri Bader		Nice, Moishe. I assume it is also proved in [2] that $\mathbb{C}^2$ and $\mathbb{P}^2(\mathbb{C})-\text{point}$ are not homeomorphic. Could the method help deciding which one is hemoe to $\mathbb{C}^2-\text{point}$ (I cannot access the paper currently)?
Mar 4, 2017 at 1:28	history	answered	Moishe Kohan	CC BY-SA 3.0