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!)