Topology – Intuition Behind Homeomorphism and Geometrical ‘Gluing Together’

continuitydefinitionfunctionsgeneral-topologyintuition

Intuitively, a homeomorphism is a way of mapping two spaces without any tearing or gluing together. Thus, I would expect the formal definition of homeomorphism in terms of continuous functions to be intuitive as well.

Specifically, the book I am reading says (translated from Chinese):

A homeomorphism is a mapping $f$ with the properties of

  1. $f$ is both surjective (don't create new points [in Chinese: 不产生新点]) and injective (without overlapping [in Chinese: 不出现重叠现象]);
  2. $f$ is continuous (without tearing [in Chinese: 不撕裂]);
  3. $f^{-1}$ is continuous (without gluing [in Chinese: 不粘连]).

The first two properties are quite intuitive to me. However, I am confused about the third one.

  1. What is gluing geometrically? What is the difference of it from non-injection (thus, overlapping) of a mapping? How is it related to the continuity of $f^{-1}$?

Consider the frequently-used example which demonstrates that the mapping
$$f: [0,1) \to S^{1}: f(t) = e^{i 2 \pi t}$$
, where $S^{1}$ is the unit circle in the complex plane,
is not a homeomorphism.
With $f^{-1}$ being not continuous, I would expect some gluing phenomenon in the deformation between $[0,1)$ and $S^{1}$.

  1. What is the gluing phenomenon in the above example?

Best Answer

Continuity of $f$ can be seen as "if you move just a bit, the image through $f$ will move a bit too".

In your case $f^{-1}$ being not continuous would imply that you can find two points on the codomain "close enough" such that their $f$-counter image are actually far away.

Think of a rectangle and glue two opposite sides together, and think of the obvious function $f$ mapping the rectangle onto this "bracelet" (suppose you restrain the domain so that $f$ is injective). If you take two points on the latter (one on one side of the glue line, the other on the other one), their $f$-counter image would be actually far from each other.

This is exactly what happens in your second question. Take the points on the circle $A=e^{i 2\pi \theta}$ and $B=e^{i 2\pi(-\theta)}$ with $0<\theta<1$ "small enough". On the circle A and B are really close, but their counter image on the interval $[0,1)$ would be $\theta$ and $1-\theta$, not really close.

Related Question