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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?2012-02-13
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    Hint: the domain of $\gamma$ should be larger for it not to be a homeomorphism.2012-02-13

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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. 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?2012-02-13
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    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...2012-02-14
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    @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 ${1\over6}$.2012-02-14
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    @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.2012-02-14
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    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?2012-02-15
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    @fatoddsun: I had remarked that your comment "Thanks a lot ..." was missing the point but didn't react then. Now I can respond: The set $U:=\bigl({1\over12},{1\over4}\bigr)$ is open in $I$, but its image $f(U)$ (a horizontal segment of length $1$ centered at $(0,0)$) is not open in $S$.2012-02-15
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    Perfect! Thank you very much!2012-02-16