Cell complex structure of real projective plane

Question

A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.

Answer 1

$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 \to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:

Two $0$-cells $d^0_\pm = \{ (\pm 1,0,0) \}$.

Two open $1$-cells $d^1_\pm = \{ (x,y,0) \mid x^2 + y^2 = 1, (-1)^{\pm 1} y > 0 \}$.

Two open $2$-cells $d^2_\pm = \{ (x,y,z) \mid x^2 + y^2 + z^2 = 1, (-1)^{\pm 1} z > 0 \}$.

Attaching maps for $d^1_\pm$ are $\phi^1_\pm : D^1 \to S^2, \phi^1_\pm(x) = \pm (\sin(\pi(x+1)/2), \cos((\pi(x+1)/2),0)$ and those for $d^2_\pm$ are $\phi^2_\pm : D^2 \to S^2, \phi^2_\pm (x,y) = \pm (x,y, \sqrt{1- x^2 - y^2})$.

For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_\pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $\psi^i$ are induced by those of the cells of $S^2$, i.e. we have $\psi^i = p\phi^i_\pm$.