# Thread: Function to map an annulus to a circle?

1. ## Function to map an annulus to a circle?

I need to describe a function which maps an annulus to its 'inner' circle.
Like, for mapping the torus X to the annulus Y, I said that for every point $x\in X$ with coordinates (i,j,k) on the torus, $f(X):x\rightarrow y| y_i=x_i, y_j=x_j, y_k=0$.. I'm looking for something similar to then map the ys onto a circle, by "contracting" it, as it were.

2. Are you wanting a nice, smooth map? A bijection? Or just some function?

I can't think of a neat one - I'm sure a neat one will exist! However, assume your annulus is centered at the origin. Then you know that you can take one closed interval [a, b] to another [c, d] by the function $x \mapsto \frac{x-a}{b-a}(d-c)+c$. So, generalise this. Take the point $(x, y)$ in your annulus and simply map it so far along the line (that is, on the positive part of the x-axis you will just be moving the interval $[a, b]$ to $[0, a]$, and just rotate this all the way around).

I will leave you to fill in the details.