Constructing image under homography from known information

Question

If $f$ is a homography of the real projective line, $f^2=id$ (is an involution), and $f$ has exactly two fixed points, how can I construct (geometrically) the image of an arbitrary point?

Answer 1

I assume you mean knowing where the fixed points are.

Assume for simplicity that the fixed points of $f$ are $0$ and $\infty$. Then it's not too hard to see (for example by the classification of Möbius transformations) that $f$ has to be $x \mapsto -x$. You know how to geometrically construct the image of an arbitrary point by that transformation.

In the general case, you can show that $f$ is the inversion with respect to the circle that contains the two points and whose center is on the real projective line.

Edit: I'm adding some details following OP's questions in the comments below.

Here I'm using the usual coordinate $x$ on the real projective line (defined by $[x_1, x_2] \mapsto x = \frac{x_1}{x_2}$) so that $\mathbb{R}\mathbf{P}^1 \approx \mathbb{R} \cup \infty$ and a homography acts like $x \mapsto \frac{ax+b}{cx+d}$.

In particular we can see homographies of the real projective line as special examples of homographies of the complex projective line $\mathbb{C}\mathbf{P}^1 \approx \mathbb{C} \cup \infty$ (aka Möbius transformations), they are those with real coefficients, or equivalently those which preserve $\mathbb{R}\mathbf{P}^1 \subset \mathbb{C}\mathbf{P}^1$.

Circles (or should I say "circle-or-line"s) are well defined in $\mathbb{C}\mathbf{P}^1$ and there is a geometric transformation called inversion with respect to a circle (see link above). I am saying that in restriction to the real projective line, a "real" homography that fixes two points on the real projective line acts like the inversion with respect to the circle (in $\mathbb{C}\mathbf{P}^1$) whose center is on the real projective line and contains the two points.