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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?
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  • $\begingroup$ 0 is being glued to the right side of the interval [0,1) $\endgroup$ Sep 21, 2014 at 14:19

1 Answer 1

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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.

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  • $\begingroup$ Nice explanation. Do you mean removing an edge of the rectangle by restraining the domain? $\endgroup$
    – hengxin
    Sep 21, 2014 at 14:46
  • $\begingroup$ Exactly. Topologically speaking you would actually have the domain being a quotient of sets, but yeah you can think of removing one edge. $\endgroup$ Sep 21, 2014 at 14:47
  • $\begingroup$ Is it OK to regard "gluing" (of $f$ on $[0,1)$ or your rectangle) as the same with "tearing" (of $f^{-1}$ on $S^{-1}$ or your bracelet)? After all, both "gluing" and "tearing" are about continuity. Briefly, the "gluing" from the perspective of $f$ is the "tearing" from the perspective of $f^{-1}$. Similarly, the "tearing" from the perspective of $f$ is the "gluing" from the perspective of $f^{-1}$. Is this understanding reasonable? $\endgroup$
    – hengxin
    Sep 22, 2014 at 1:26
  • $\begingroup$ Yeah, it's legit. As long as $f$ is bijective, you can see it in both points of view: you "glue" the rectangle to a bracelet, and you "rip" the bracelet to a rectangle, it's just a question of domains and codomains. $\endgroup$ Sep 22, 2014 at 12:28

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