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In the projective space $\mathbb{P}^n(\mathbb{K})$ I can consider the hyerplane $H_{0}=\{x_{0}=0\}$ and the set $U_{0}=\mathbb{P}^n(\mathbb{K})-H_{0}$. Clearly $\mathbb{P}^n(\mathbb{K})=U_{0}\cup H_{0}$. The functions

  • $j_{0}:\mathbb{K}^n\to U_{0}$ defined by $j_{0}(x_{1},...,x_{n})=[1,x_{1},...,x_{n}]$,
  • $i_{0}:H_{0}\to \mathbb{P}^{n-1}(\mathbb{K})$ defined by $i_{0}([0,x_{1},...,x_{n}])=[x_{1},...,x_{n}]$.

Are obviously both bijections. My question is: are them also homeomorphisms? Why?

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    If the topologies involved are the canonical topologies (as I mentioned in the last comment), you are right, they are homeomorphisms. Think about how are you defining the topology in $\mathbb P^n(\mathbb R)$.2012-01-29

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Hint about why $j_0$ is continuous: $U_0$ is open in $\mathbb P^n(\mathbb R)$ because $ \pi^{-1}(U_0)=\{x=(x_0,x_1,\dots,x_n)\in\mathbb R^{n+1}-\{0\}: x_0\neq0\} $ is open in $\mathbb R^{n+1}-\{0\}$.

Let $V$ an open set in $\mathbb P^n(\mathbb R)$. You should find what is the relation between $j_0^{-1}(V)$ and $\pi^{-1}(V)$, and then prove that $j_0(V)$ is open.

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    Hola! :) ${}{}{}$2012-01-31