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Consider the map $$\gamma\colon (0,1)\to\mathbb{R}^2,\ t\mapsto (\cos(2\pi t),\sin(2\pi t)).$$ This is an example of a map which is continuous and injective but not a homeomorphism onto the image, since the inverse could not be continuous. In fact, two points arbitrarily close to each other in a small neighbourhood of $(1,0)$ would go far apart in the preimage. By definition, a function is continuous if the preimage of every open set is open in the domain. How could I find an open set in the support of this curve which is sent to a non-open set in the interval?

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"How could I find an open set...": you can't, because $\gamma$ is a homeomorphism onto its image. Can you see your fallacy? – Georges Elencwajg Feb 13 '12 at 12:07
Hint: the domain of $\gamma$ should be larger for it not to be a homeomorphism. – Ilya Feb 13 '12 at 12:22
up vote 11 down vote accepted

The given map is a homeomorphism onto the image: Let $I:=(0,1)$, $S:=\gamma(I)$, and consider a point $z\in S$. Then $t:=\gamma^{-1}(z)\in I$. Any neighborhood $V$ of $t$ contains an open interval $(a,b)$ such that $0<a<t<b<1$. The set $$\Omega:=\{(x,y)\in{\mathbb R}^2\ |\ x^2+y^2>0, \ 2\pi a<\arg(x,y)<2\pi b\}$$ is open in ${\mathbb R}^2$, whence $U:=\Omega\cap S$ is an open subset of $S$ which contains the point $z$. Therefore $U$ is an open neighborhood of $z$, and $\gamma^{-1}(U)=(a,b)\subset V$.

Here is an example of a continuous injective map $f:\ I\to {\mathbb R}^2$ which is not a homeomorphism onto its image: $$f(t)\ :=\ \cases{(6t-1,0) & $\bigl(0 < t\leq{1\over3}\bigr)$\cr (2-3t, 3t-1) & $\bigl({1\over3}\leq t\leq{2\over3}\bigr)$ \cr (0,3-3t) & $\bigl({2\over3}\leq t < 1\bigr)$ \cr}\quad.$$ Drawing a figure one sees that the inverse map $f^{-1}$ is not continuous at $(0,0)=f\bigl({1\over6}\bigr)$.


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Wow: so I was completely wrong from the very beginning! Thanks a lot for pointing this out! So... can you find an example of a map $(0,1)\to\mathbb{R}^2$ continuous and injective which is not a homeomorphism onto the image? – fatoddsun Feb 13 '12 at 14:00
Thanks a lot... but I still have to find an open set in $S$ such that its preimage in $I$ is not open. I guess I should look at the origin.... but I can't find it yet... – fatoddsun Feb 14 '12 at 17:25
@fatoddsun, concerning my example: $f^{-1}(0,0)={1\over6}$, but for any open set $\Omega\subset{\mathbb R}^2$ containing $(0,0)$, however small, the set $f^{-1}(S\cap\Omega)$ contains points close to $1$, whence far away from ${1\over6}$. – Christian Blatter Feb 14 '12 at 18:44
@fatoddsun You cannot. The invariance of domain theorem says that any continuous and injective map $f$ defined on an open subset $O$ of any $\mathbb{R}^n$ with image in some $\mathbb{R}^m$ will be a homeomorphism between $O$ and $f[O]$. Full invariance of domain is probably overkill for the case $n=1, m=2$ that you have here, a connectedness argument of sorts should probably do. – Henno Brandsma Feb 14 '12 at 20:02
Wait, wait, wait! I think we (me, in particular...) are getting a bit confused! We are saying that the map $f$ defined by Christian Blatter is NOT a homeomorphism onto the image, but it still a continuous map, isn't it? Hence I was wrong asking for an open set in $f(I)$ whose preimage was not open as this is the definition of continuity! What I'd like to find is an open set in $I$ which is mapped to a non-open set in $f(I)$, and this must exist, otherwise $f$ would be open, thus a homeomorphism onto the image. Am I right, now? – fatoddsun Feb 15 '12 at 13:48

Maybe it is too late to ask this question, but while going through this example, I was wondering whether the function suggested by Christian Blatter, while not a homeomorphism from $(0, 1) \mapsto {\mathbb R}^2, $ is still an embedding from $(0, 1) \mapsto f((0, 1)) $ once $f((0, 1)) $ is equipped with the relative topology of $\mathbb{R}^2 $ intersected with $f((0, 1)) $?


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This probably would be better posted as a new question. Questions asking for clarifying some part of an answer to another question are valid questions; see Clarify an old answer on meta. (You will probably agree that your post does not answer the question, so it is not an ideal thing to be posted in the answer field.) – Martin Sleziak Oct 27 '15 at 5:59
Other possibility would be to post a comment on the answer you want to ask about. (You have to judge for yourself which possibility is better.) – Martin Sleziak Oct 27 '15 at 6:00

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