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.
Best Answer
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.