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Two descriptions of projective spaces are same.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Definition.Identification topology: Let $X$ be a topological space and let $\mathcal{P}$ be a partition of $X$. Consider a set $Y$ whose elements are the members of $\mathcal{P}$. Define the function $\pi : X \rightarrow Y$ which sends each point of $X$ to the subset of $\mathcal{P}$ containing it. A subset $O \subseteq Y$ is open in $Y$ if and only if $\pi ^{-1}(O)$ is open in $X$. So this gives a topology to $Y$ known as the identification topology on Y.

Projective space description 1:
Consider the unit circle $S^1$ in $\mathbb{R}^2$ and partition into subsets which contain exactly two points, the points being antipodal (at the opposite ends of the diameter). $P^2$ is the resulting identification space.

Projective space description 2:
Begin with $\mathbb{R}^2 - \{ 0 \}$ and identify two points if and only if they lie on the same straight line through the origin. The resulting identification space is $P^2$.

Show that the two descriptions of $P^2$ are equivalent.

I have no idea how to begin. Please help.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
You need to start by constructing a bijection between PS1 and PS2. For example, given a line $L$ through the origin in $\mathbb R^2$, you could define $f(L) = L\cap S^1$ (so that $f(L)$ is a pair of antipodal points in $S^1$). That gives the required bijection. You then need to check that this map takes the topology on PS2 to the topology on PS1. To do that, you must show that if $U$ is the union of a set of lines through the origin which is open as a subset of $\mathbb R^2$, then $f(U) = U\cap S^1$ is an open subset of $S^1$, and conversely.