# projective plane is homoemorphic to the circle

Let $S^{1}$ be a unit circle in $R^{2}$,and let $X$ be the space obtained from $S^{1}$ by identifying antipodal points. Show that projective space $X$ is homeomorphic to $S^{1}$

First observation is that the half circle H={ $e^{\theta i}$ where $0\leq {\theta} \leq \pi$} by identifying the antipodal points will be homeomorphic to the projective space for $S^{1}$.

Second it's enough for us to prove that $H/ \sim$ is homeomorphic to the $S^{1}$. Then consider the map $f=z^{2}$ from H to $S^{1}$. Since the value of function is the same for the same equivalent class of $H$, the we can induce a bijective map from $H/\sim$ to $S^{1}$ by $f$,and the continuity can be obtained from universal property of quotient map. Then $H/\sim$ is Hausdorff and $S^{1}$ is compact then it's homeomorphism.

My question is that how to prove the first observation that $H/\sim$ is homeomorphic to $S^{1}/\sim$ by rigorous arhuments including the bijective and continuity from both sides.

• The terminology of your title is incorrect. $X$ is called the "projective line", not the "projective plane". The projective plane is something different, obtained from $S^2$ by identifying antipodal points. – Lee Mosher Jan 13 '17 at 16:24
• Your paragraph starting "Second it's enough..." is a rigorous proof, so it is unclear what you are asking. – Lee Mosher Jan 13 '17 at 16:26