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Briefly speaking, we know that a map $f$ between $2$ topological spaces is homeomorphic if $f$ is a bijection and the inverse of $f$ and itself are both continuous.

So, can anyone give me $2$ counter examples(preferably simple ones) of non-homeomorphic maps $f$ between 2 topological spaces that satisfy the properties I give? (Only one of them is satisfied and 3 examples for each property.)

  1. $f$ is bijective and continuous, but its inverse is not continuous.
  2. $f$ is bijective and the inverse is continuous, but $f$ itself is not continuous.

In addition, can we think about some examples of topologies that are path-connected?

I will understand the concept of homeomorphism much better if I know some simple counterexamples. I hope you can help me out. Thanks!

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  • $\begingroup$ Clearly if $f$ and $f^{-1}$ are an example for #1, then $f^{-1}, f$, in that order is an example for #2. For #3, no example is possible: a function with an inverse is necessarily bijective. $\endgroup$ – rschwieb Apr 15 '13 at 17:49
  • $\begingroup$ Sorry, my bad again. What I mean here is that only one property is satisfied. :) $\endgroup$ – Cancan Apr 15 '13 at 17:50
  • $\begingroup$ Related. $\endgroup$ – Cameron Buie Apr 15 '13 at 18:28
  • $\begingroup$ here are examples where you have a continuous bijection $f: X \to Y$ and a continuous bijection $g: Y \to X$ but $X$ and $Y$ are not homoemorphic. $\endgroup$ – s.harp Feb 6 '16 at 11:07
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Consider $f:[0,2\pi)\to S^1$ given by $t\mapsto\langle\cos t,\sin t\rangle,$ where $S^1$ is the unit circle in the plane, and $[0,2\pi)$ is the real interval, both considered under their appropriate subspace topology as subsets of Euclidean spaces. Then $f$ is bijective and continuous, but its inverse is not continuous, providing an example for 1. Thus, its inverse is an example for 2.

Be careful with 3, as you need to specify what you mean by "inverse" in cases where $f$ isn't bijective. An equivalent way to define homeomorphism is as a bijective, continuous, open map (maps open sets to open sets). This avoids the need to worry about inverse functions--indeed, open maps need not be injective. For an example of a surjective, continuous, open map that is not a homeomorphism (since it is not injective), consider $p:\Bbb R^2\to\Bbb R$ given by $\langle x,y\rangle\mapsto x$, where $\Bbb R^2$ and $\Bbb R$ are in their usual topologies. For an example of an injective, continuous, open map that is not a homeomorphism (since it is not surjective), consider the inclusion $(0,1)\hookrightarrow[0,1].$

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  • $\begingroup$ But who will you prove that the inverse is not continuous(with delta elsilon definition maybe)? $\endgroup$ – Cancan Apr 15 '13 at 19:19
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    $\begingroup$ There are several ways to go about it. If you know that continuous functions map compact sets to compact sets, and that "compact"="closed and bounded" for subsets of Euclidean spaces, then we may simply observe that $S^1$ is compact, while $[0,2\pi)$ isn't. If you aren't familiar with that, try taking the limit of $f^{-1}(\langle x,y\rangle)$ as $\langle x,y\rangle$ approaches $\langle 1,0\rangle$ from a clockwise direction and as $\langle x,y\rangle$ approaches $\langle 1,0\rangle$ from a counterclockwise direction. For $f^{-1}$ to be continuous, those limits must be the same. $\endgroup$ – Cameron Buie Apr 15 '13 at 19:30
  • $\begingroup$ Sorry, you are right! I didn't notice it's open at $2\pi$ $\endgroup$ – Cancan Apr 15 '13 at 21:10
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$1.$ Let $X$ be the set of real numbers, with the discrete topology, and let $Y$ be the reals, with the ordinary topology. Let $f(x)=x$. Then $f$ is continuous, since every subset of $X$ is open. But $f^{-1}$ is not continuous.

$2.$ In doing $1$, we have basically done $2$.

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  • $\begingroup$ You must have posted the answer while I was typing up mine. It is almost identical to yours. Sorry I did not see your answer. $\endgroup$ – Rankeya Apr 15 '13 at 17:54
  • $\begingroup$ Haha, interesting, but both of you are correct for the first 2 properties. but how about the 3rd one? any idea? $\endgroup$ – Cancan Apr 15 '13 at 17:57
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    $\begingroup$ Dear @Cancan: It does not make sense to talk of the inverse of a function which is not a bijection. By saying that the inverse of a function exists, you are also saying that it is a bijection. So, for 3 you are really asking us to give you an example of a homeomorphism. $\endgroup$ – Rankeya Apr 15 '13 at 17:58
  • $\begingroup$ The third one is problematic, since by the usual definition there is no inverse. $\endgroup$ – André Nicolas Apr 15 '13 at 17:59
  • $\begingroup$ @Rankeya ,Andre You guys are right. I thought it wrong before. but can you guys give examples of path connected topology rather than discrete topology? $\endgroup$ – Cancan Apr 15 '13 at 18:00
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1 and 2 are really the same thing. Also, what do you mean by the inverse of a function if it is not a bijection (your question 3)? The inverse of a function is not defined for something that is not bijective.

For an example of 1 take $X = \mathbb{Q}$ with the discrete topology and $Y = \mathbb{Q}$ with the usual topology inherited from $\mathbb{R}$. Let $Id: X \rightarrow Y$ be the identity map on the level of sets. This map is clearly continuous and a bijection. But, it is not open hence the inverse is not continuous.

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$f:[0,2\pi)\to S^1$ given by $f(t)=(\cos(t),\sin(t))$ is continuous and bijective but its inverse is not continuous.

The inverse of $f$ is an example of a bijection $f^{-1}:S^1\to [0,2\pi)$ that is not continuous, but its inverse is continuous.

It is very illustrative to think geometrically about this example. It just involves gluing the interval $[0,2\pi)$ to get the circle.

If $f$ has an inverse then $f$ is a bijection.

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Another beautiful counter-example is perhaps $id:\left(\mathbb{R},usual\right)\to \left(\mathbb{R},Cof\right)$ $\ \ $ where $\ Cof\ $ is the topology for which every open set has a finite complement. Clearly $id$ is continuous (inverse image of an open set in $Cof$ is union of intervals) and bijective, however $id^{-1}$ is not continuous because $id^{-1}[(a,b)]=(a,b)\not\in Cof$. Note that in this example $Y$ is not Hausdorff.

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