# $(0,1)\to\mathbb{R}^2$ injective, continuous, not a homeomorphism on the image

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

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 $(0,0)=f\bigl({1\over6}\bigr)$.
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