Show that the real projective line is homemorphic to the unit circle 
Define the unit sphere as $S^1=\{x\in \mathbb{R}^2: \|x\|=1\}$
Also define the real projective line as $\mathbb{R}P^1=S^1/(x\sim-x)$

We can consider the mapping $f:S^1\rightarrow S^1$, $f(x)=x^2$
If I can show that $f$ is a continuous quotient map, i.e. $f$ is a continuous surjective mapping such that $f(-x)=f(x)$ for all $x\in S^1$, then I can apply the universal property of a quotient topology and conclude that there exists an induced homeomorphism between $\mathbb{R}P^1$ and $S^1$.
I am unsure how to prove, however, that $f$ is both surjective and continuous. It's obvious that $f(-x)=f(x)$ for all $x\in S^1$, but how should I go about the other two claims? I think I am overthinking this. Any help would be much appreciated.
 A: The reason Ashvin gave for surjectivity, using the polar coordinates representation is perfectly fine:

The map $f$ is surjective because in polar coordinates, it is given by $e^{i\theta} \mapsto e^{2i\theta}$, and every angle $\psi \in [0, 2\pi)$ can be uniquely represented as $2\theta$ for some $\theta \in [0, \pi)$.

(Or simply note that all polynomials are surjective in $\mathbb{C}$ by the universal theorem of algebra and $|z^2| = |z|^2$ so the norm of a preimage of a point of norm $1$ is still $1$.)
Continuity is in fact easy: it follows from the universal property of quotient maps : if $q: X \to Y$ is a quotient map (So $Y$ has the quotient topology wrt $q$ and $X$) then $g: Y \to Z$ is continuous iff $g \circ q: X \to Z$ is continuous. So continuity from a quotient space is determined by the composition with the quotient map.
In our case $X = S^1$ and $q: S^1 \to \mathbb{R}P^1$ is given by $q(x) = \{x,-x\}= [x]$, the class of $x$ under $\sim$. 
The map $f(z) = z^2$ has the property that $f(z) = f(x)$ iff $z^2 = x^2$ (squares taken as members of $\mathbb{C}$) iff $x^2 - z^2 = 0$ iff $(x-z)(x+z) = 0$ iff $x=z$ or $x=-z$ iff $x \sim z$.
The fact that $x \sim x'$ implies $f(x) = f(x')$ implies that the map $\overline{f}: \mathbb{R}P^1 \to S^1$ defined by $\overline{f}([x]) = f(x)$ is well-defined: for all representatives of the class of $x$, ($x$ and $-x$), we get the same $f$-value. But then this induced map $\overline{f}$ then obeys by definition
$$\overline{f} \circ q = f$$ and $f$ is continuous, so $\overline{f}$ is continuous by the aforementioned universal property directly.
The property that $f(x) = f(x')$ implies $x \sim x'$ implies that $\overline{f}$ is 1-1: $\overline{f}[x] = \overline{f}[x']$ iff $f(x) = f(x')$ iff $[x] = [x']$.
So $\overline{f}$ is a 1-1 continuous and onto (if $z \in S^1$ find $x\in S^1$ with $f(x) = z$ and then $\overline{f}([x]) = z$ too) map from the compact space $\mathbb{R}P^1$ (continuous image of $S^1$ under $q$) to $S^1$, hence a homeomorphism as $S^1$ is Hausdorff.
A: The map $f$ is surjective because in polar coordinates, it is given by $e^{i\theta} \mapsto e^{2i\theta}$, and every angle $\psi \in [0, 2\pi)$ can be uniquely represented as $2\theta$ for some angle $\theta \in [0, \pi)$. The map $f$ is a differentiable function from $\mathbb{R}$ to $S^1$, and it is periodic with period $2\pi$, so it is continuous as a function from $\mathbb{R}/(2\pi \mathbb{Z}) \simeq S^1$ to $S^1$.
