Show that $\mathbb{R}P^1 \simeq S^1$ I have known for a while that $\mathbb{R}P^1 \simeq S^1$. Indeed, visually it is easy to see why this is the case (any two antipodal points define a line through the origin)
Say we now forget everything we know, and let $F$ be a division ring and let $n \ge 0$. We definie an equivalence relation of $F - \{ 0 \}$ by $x \sim y$ if there exists $\lambda \in F - \{ 0 \}$ with $x = \lambda y$. The quotient set $F^{n+1}-\{ 0 \} / \sim$ is called the F-projective n-space. The class of $x = (x_0,\ldots ,x_n)$ is denoted by $[x] = [x_0, \ldots, x_n] \in FP^n$.
Based on this definition I would like to show $\mathbb{R}P^1 \simeq S^1$
So how do we go from this definition to prove the isomorphism? I am sure it is simple...but I am not really sure where to go. Do I just use the antipodal max $f(x)=-x$? (Note the next part is to extent to show $\mathbb{H}P^1 \simeq S^4$ which is a bit harder to visualise!)
 A: First, consider the mapping
$$\varphi\colon (x,y) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}\right)$$
which maps $\mathbb{R}^2-\{0\}$ to the unit circle $\mathcal{C}$.
If $(x,y)$ and $(z,w)$ map to the same point, then it follows that $x = \lambda z$ and $y=\lambda w$, where 
$$\lambda = \frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}.$$
However, it is possible for $(x,y)$ and $(z,w)$ to satisfy $(x,y)=\lambda(z,w)$ and not to map to the same point: indeed, if $(x,y)=\lambda(z,w)$, then it is easy to see that $(x,y)$ maps to
$$\frac{\lambda}{|\lambda|}\left(\frac{z}{\sqrt{z^2+w^2}},\frac{w}{\sqrt{z^2+w^2}}\right).$$
So we define an equivalence relation on the unit circle by $\mathbf{x}\sim\mathbf{y}$ if and only if $\mathbf{x}=\pm\mathbf{y}$, and consider $\mathcal{C}^1/\sim$. 
Now we consider the composition of $\varphi$ and $\pi\colon \mathcal{C}^1\to\mathcal{C}^1/\sim$. If $(x,y) =\lambda (z,w)$, then $\pi\circ\varphi(x,y) = \pi\circ\varphi(z,w)$ by the discussion above. Conversely, if $\pi\circ\varphi(x,y) = \pi\circ\varphi(z,w)$, then either $\varphi(x,y)=\varphi(z,w)$, in which case $(x,y) = \lambda(z,w)$ with $\lambda$ as above; or else $\varphi(x,y) = -\varphi(z,w)$, hence
$$(x,y) = -\frac{\sqrt{x^2+y^2}}{\sqrt{z^2+w^2}}(z,w).$$
So we conclude that $\pi\circ\varphi$ factors through $\mathbb{R}^2-\{(0,0)\}/\equiv$, where $\equiv$ is the equivalence relation on $\mathbb{R}^2=\{(0,0)\}$ given by $\mathbf{x}\equiv \mathbf{z}\Longleftrightarrow \mathbf{x}=\lambda\mathbf{z}$ for some scalar $\lambda$, and induces a bijection
$$\frac{\mathbb{R}^2-\{(0,0)\}}{\equiv} \to \mathcal{C}^1/\sim \cong S^1.$$
The maps are all continuous.
A: I suppose you want a homeomorphism between $\mathbb R\mathbb P^1$ and $S^1$. ($S^1$ does not have an algebraic structure, so isomorphism in that sense would not make sense.)
Define $\mathbb R\mathbb P^1\to S^1$ by $[a:b]\mapsto (a/\sqrt{a^2+b^2}, b/\sqrt{a^2+b^2})$ and $S^1\to \mathbb R\mathbb P^1$ by $(a,b)\mapsto [a:b]$. These are inverses to each other and hence bijective. They are both clearly continuous, so we're done.
ps: Why do you skip $\mathbb C\mathbb P^1\sim S^2$?
pps: the proof is the same in all cases including $\mathbb C$ and $\mathbb H$. You just have to figure out the dimension of each division algebra over $\mathbb R$, but it seems you already know that.
A: Just to have in mind a picture of the procedure of identifying opposite points on the circle, you can think on this:

