What is the mapping torus of f(theta)=2*theta on S^1 Let $X=S^1$ and $f:X\to X$.
If $f$ is defined by $f(\theta)=2\theta$, what does the mapping torus $T_f$ look like?
(A picture would be greatly appreciated.)
I can easily imagine $T_f$ when $f$ is an injective function, e.g. $f(\theta) = -\theta$.
But when $f(\theta)=2\theta$, then $T_f = X\times [0,1]\ /\ \{(x,0)\sim (f(x),1)\}$ does not contain a copy of $X$ at time 0 or 1.
Do we normally only consider the mapping torus $T_f$ when $f$ is a bijection?
 A: I would first visualize the mapping cylinder $M_g$ of $g: [0,1] \rightarrow S^1 $, where we see $S^1$ as $[0,1]$ with $0$ and $1$ identified and $g$ is the quotient. This mapping cylinder looks like a cone, but instead of a point, we have a line at the top. Try to draw this!
Then consider the mapping cylinder $M_{gr}$ of $g$ composed with he reflection $r: [0,1] \rightarrow S^1: t \mapsto 1-t$. ($M_{gr}$ is the reflection of the first figure) Identify $M_g$ and $M_{gr}$ at the extreme points of the upper line, that is, $(0,0) \in M_g \sim (0,0) \in M_{gr}$ and $(1,0) \in M_g \sim (1,0) \in M_{gr}$, and then you also identify the lower circles, that is, $(t, 1) \in M_g \sim (t, 1) \in M_{gr}$.
If you look to this picture in $\mathbb{R}^3$, you will see selfintersections, but these intersections doesn't occur. (The problem is that in $\mathbb{R}^3$ you can't embed this construction. Treat these intersections as the selfintersections of the Klein bottle when you see it in $\mathbb{R}^3$, ie. ignore them)
Finally, identify the upper and the lower circles in the obvious way in order to obtain the mapping torus you were looking for.
