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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.

Jack
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  • 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
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    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

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Assuming you want to prove that the unit circle is homeomorphic to the real line. Let $S^1 \subset \mathbb{R}^2$ be the unit circle, and let $N = (0,1) \in S^1$ be the north pole. Now, we need to find a homeomorphism between these two spaces. Consider the stereographic projection

$$ f : S^1 - \{N\} \xrightarrow{\cong} \mathbb{R} $$ mapping a point $(x,y) \in S^1 - \{N\}$ to $t \in \mathbb{R}$, where $(t, 0)$ is the intersection of the line through $N$ and $(x,y)$ with the $x$-axis. Hence $$ f(x,y) = t = \frac{x}{1 - y} $$ for all $(x, y) \in S^1 - \{N\}$. This is the restriction to the subspace $S^1 - \{N\}$ of a continuous function from $\mathbb{R} \times (\mathbb{R} - \{1\}) \subset \mathbb{R}^2$ to $\mathbb{R}$, hence $f$ is continuous.

The inverse function $f^{-1}: \mathbb{R} \to S^1 - \{N\}$ maps $t$ to $f^{-1}(t) = (x, y)$ with $t = \frac{x}{1 - y}$ and $x^2 + y^2 = 1$, which we can solve for $y \neq 1$ and then $x$ to obtain $$ f^{-1}(t) = (x, y) = \left( \frac{2t}{t^2 + 1}, \frac{t^2 - 1}{t^2 + 1} \right). $$

Clearly $f$ and $f^{-1}$ are continuous and are inverse homomorphisms. This result can be generilzed to $n$ dimensions by the way.

Nikola Golis
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