Prove that $\mathbb R\mathrm P^1$ is diffeomorphic to certain submanifold of $\mathbb R^3$.

Solution 1:

$\textbf{Claim:}$ The image $f(\mathbb{R}P^1)$ is a circle of radius $\frac{1}{2}$ lying in the plane $u+w = 1$ with center $\left(\frac{1}{2},0\right)$.

We will denote the coordinates on $\mathbb{R}^3$ as $(u,v,w)$ as suggested by Thomas in the comments, and derive the suggested relations. Now $u(x:y) = \frac{x^2}{x^2+y^2}$ and $w(x:y) = \frac{y^2}{x^2+y^2}$. Hence $u(x:y) + w(x:y) = 1$ for all $(x:y)\in \mathbb{R}P^1$. So, the image of $f$ has to be contained in the plane $u+w = 1$. A similar calculation verifies that the image also satisfies $v^2 = uw$.

Now rearranging the equation for the plane we get $w = 1-u$. Replacing $w$ in the equation $v^2 = uw$ and rearranging, we get, $$v^2 +u^2 - u = 0$$ Finally, by completing the square we arrive at $$v^2 +\left(u-\frac{1}{2}\right)^2 = \frac{1}{4} $$