I know intuitively that the mapping given by $$ p : \mathbb{R}^2 \to \{[(x,y,1)]\ |\ x,y \in \mathbb{R}\}\subset \mathbb{P}_2(\mathbb{R}), p((x,y)) = [(x,y,1)] $$ will be an embedding from ${\mathbb{R}^2}$ to ${\{[(x,y,1)]\ |\ x,y \in \mathbb{R}\}\subset \mathbb{P}_2(\mathbb{R})}$. But how would I show it formally? It's clearly bijective, but I need to show continuity of $p$ and it's inverse, which involves looking at the open subsets of ${\{[(x,y,1)]\ |\ x,y \in \mathbb{R}\}}$, for which I am a bit stuck
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1Perhaps the answerer has a "point"! – amWhy Oct 17 '21 at 20:58
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@amWhy is that a clue? – Riemann'sPointyNose Oct 17 '21 at 21:00
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1Maybe a "point", maybe a "carrot". – amWhy Oct 17 '21 at 21:01
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You can use the fact that $\mathbb{RP}^2$ is a quotient of $\mathbb R^3-{0}.$ In particular, a set in $\mathbb{RP}^2$ is open if and only if its preimage in $\mathbb R^3$ is open. – D. Brogan Oct 17 '21 at 21:06