# Projective space as a manifold

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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The answer is in the topologies involved. What are they? –  Davide Giraudo Jan 29 '12 at 15:11
I think he is thinking in $\mathbb K=\mathbb R$ or $\mathbb C$, and the canonical topologies (induced by $\mathbb R^n$ as a metric space with the distance $d(x,y)=\|x-y\|$). He haven't asked about algebraic geometry before, so I don't think that is the Zariski topology. –  emiliocba Jan 29 '12 at 15:23
Yeah right emiliocba anticipate me. I'm thinking the Euclidean topology in $\mathbb{R}^n$ and the Final topology (the finest topology such that $\pi:(\mathbb{R}^{n+1}-\{0\})\to \mathbb{P}^n(\mathbb{R})$ is continuous.) –  Lorenzo Rossi Jan 29 '12 at 15:26
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 $\mathbb P^n(\mathbb R)$. –  emiliocba Jan 29 '12 at 15:27

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.
Hola! :) ${}{}{}$ –  Mariano Suárez-Alvarez Jan 31 '12 at 7:06