# Homeomorphism between the 1-sphere and a semi-open real interval

I need help with a problem that's troubling me.

In Lee's "Introduction to Topological Manifolds" I found this exercise: being given the exponential map $\ a:[0,1[\to\mathbb{S}^{1}$, $\ a(s)=e^{2\pi i s}\$ from $\ [0,1[\subseteq\mathbb{R}\$ to the unit 1-sphere in $\mathbb{C}$, both with the metric topology (of $\mathbb{R}$ and $\mathbb{C}$), show that it is continuous and bijective, but not a homeomorphism. First of all I tried to give the most straight forward proof, but it seems so trivial to me that I'm not sure whether it is correct or not, so I would like to know if you think that it is suitable. Then I went further and changed the topologies given by the problem (of course, not expecting to prove the same result), and I got a weird result! I found the 1-sphere to be homeomorphic to the semi-open interval, which I know by sure that can't be possible. I'd like to know where I went wrong. I know that there may be other ways to prove the non-homeomorphicity, but I'm interested in the results that I got with these ones.

This is the first proof, following Lee's requests: $\ a$ is in fact bijective, with $s(a)=(2\pi i)^{-1}\ \ln a\$ as its inverse. However, $\mathbb{S}^{1}$ is not open nor closed in the metric topology of $\mathbb{C}$, not being the union or intersection of any open or closed ball in $\mathbb{C}$, neither is open or closed in $\mathbb{C}$ any subset of the 1-sphere. So I'm left with $a^{-1}(\varnothing)=\varnothing$, open preimage of the only open subset of the unit 1-sphere. So $a$ is continuous. Given any open interval $\ ]a,b[\subseteq[0,1[$, its preimage through $\ a^{-1}$ is a subset of the 1-sphere, which again is not open, so $a^{1}$ is not continuous, and $a$ is not a homeomorphism. This proof, however, does not use any information from the specific form of the map to prove its continuity, and this is the reason why I didn't appreciate it.

As I recognized that in the topologies specified by the exercise no non-trivial open subset of the 1-sphere could be used to prove by definition the continuity of $a$, I changed the topologies, and tried to give a second proof of the non-homeomorphicity between the semi-open interval and the 1-sphere. So I gave $[0,1[$ and $\mathbb{S}^{1}$ the subspace topology. Now both $[0,1[$ and $\mathbb{S}^{1}$ are open in their respective topologies, and there exist open subsets of the 1-sphere. I still used the exponential map. Obviously $a^{-1}(\mathbb{S}^{1})\$ and $a^{-1}(U)\$ are open for every $U$ not cointaining $1$. If otherwise $V\ni 1$, $a^{-1}(V)=[0,a[\cup]b,1[\$ for some $a,b<1$, so its pre-image is still open. Then $a$ is continuous (and, again, bijective by the same inverse as before). But now $a([0,1[)\$ is open, and $a(]a,b[)$ is open for every $0<a<b<1$, so $a^{-1}$ is continuous and $a$ should be a homeomorphism, proving that the semi-open interval and the 1-sphere are homeomorphic!

The only reason by which I can explain this last result is that the exponential map is a quotient map, so I really have proved the homeomorphicity between the circle and the quotient of the interval. But I didn't intend to do so, nor i defined explicitly such a quotient on the full interval $[0,1]$. So I would like to know what is the theoretical reason for this seemingly existing homeomorphism, and if I went wrong somewhere.

EDIT: Ok, I found what's wrong in the second proof. I didn't take into account the intervals of the form $[0,b[$, $b<1$, which are open in $[0,1[$ with the subspace topology, but have a not open pre-image by $a^{-1}$. So, again, $a$ is continuous and bijective, but not a homeomorphism. Nevertheless, let me know if you find the proofs correct, I'm a starter and self-learner in topology and would appreciate any advice.

Your first "proof" is senseless because you are supposed to take the subspace topology on $\mathbb{S}^1$ and $[0, 1[$ (and by the way, $\mathbb{S}^{1}$ is closed in $\mathbb{C}$!).
You are correct that, for $b < 1$, $a([0, b[)$ isn't open. However, I believe you should argue this in more details.
Lastly, the usual proof of non-homeomorphicity of $\mathbb{S}^{1}$ and $[0, 1[$ involves noticing $\mathbb{S}^{1}$ is compact while $[0, 1[$ isn't.
• You are correct about the closedness of $\mathbb{S}^{1}$ in $\mathbb{C}$, I totally overlooked that! However, Lee wasn't clear, as he said to prove it "in the metric topology of $\mathbb{R}$ and $\mathbb{C}$" (and that is why I then thought I had to shift to the subspace topology). Are you then suggesting that homeomorphisms are to be defined only between topological spaces, and not on subsets of topological spaces, unless these are intended to be themselves topological spaces with the subspace topology (so it was obvious that I had to use the subspace topologies)? – Giorgio Comitini Apr 27 '15 at 11:17
• As for the second proof, what details do you think are missing? Then, about the compactness, I know that is a proof, but I need specifically the "by definition" proof through the map $a$. – Giorgio Comitini Apr 27 '15 at 11:22
• Re: your second question, you should prove that $a([0, b[)$ isn't open – Pedro Apr 28 '15 at 11:44
• Well, that's by the definition of the function $a$, isn't it? I mean, does it require anything else to be said? – Giorgio Comitini Apr 28 '15 at 12:31