# Cell complex structure of real projective plane

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.

$$RP^2$$ 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 $$\psi^i = p\phi^i_\pm$$.