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Reading this post here Problem in proving that $\mathbb{A}^2$ is not homeomorphic to $\mathbb{P}^2$

I came up with the following question:

Why are the $\mathbb A^1$ and $\mathbb P ^1 $ homeomorphic? (with the Zariski topology endowed, and algebraically closed underlying field)

Sorry, if this is something very easy, but I can't understand it.

Any help would be really appreciated!

Community
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passenger
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1 Answers1

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Since an algebraically closed field is necessarily infinite, $\mathbb{A}^1$ and $\mathbb{P}^1$ have the same cardinality, and on both the Zariski topology is the cofinite topology. Therefore they are homeomorphic.

Sal
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  • Thank you very much for your immediately reply. Could you please give some details why cofiniteness implis that they are homeomorphic? Can we find an explicity homeomorphism between them? – passenger Dec 04 '14 at 00:08
  • Any bijection between them is a homeomorphism because bijections send cofinite sets to cofinite sets. – Sal Dec 04 '14 at 00:22
  • Could you give an example of such a bijection? I can't understand where the $\infty$ is mapped to the affine line. – passenger Dec 04 '14 at 00:24
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    Any sets of the same cardinality have a bijection between them (that is the definition of having the same cardinality). There is no natural choice of bijection but one way of making one is to choose an arbitrary sequence of distinct elements $S={s_1,s_2,\ldots}\subset\mathbb{A}^1$ and then map $\mathbb{P}^1$ to $\mathbb{A}^1$ by $$f(t)=\begin{cases} t & \text{if }t\notin S\ s_1 & \text{if }t=\infty\ s_{i+1} & \text{if }t=s_i \end{cases}$$ – Sal Dec 04 '14 at 00:27
  • (That is a standard sort of trick for bijections between an infinite set and a subset. There are plenty of threads on this site where it's used, such as this one and this one.) – Sal Dec 04 '14 at 00:30
  • Thank you! One last question, just to be sure: the continuity is follows from the fact that the Zariski topology is cofinite in both, right? – passenger Dec 04 '14 at 00:31
  • Yes. If $f:X\to Y$ is a bijection and $A\subset X$ and $B\subset Y$ are cofinite, then $f(A)$ is cofinite in $Y$ and $f^{-1}(B)$ is cofinite in $X$ since the sets $f(X\setminus A)$ and $f^{-1}(Y\setminus B)$ are finite. If the topologies on $X$ and $Y$ are defined so that cofinite sets are open, this is saying that $f$ and $f^{-1}$ are continuous. – Sal Dec 04 '14 at 00:39
  • Everything is clear to me now. Thank you very much for your time and the detailed explanations! – passenger Dec 04 '14 at 00:41