Is e: [0,1] --> S^1 a Quotient Map?

[PsychonautQQ]

Hey PF!
As usual, I'm having issues understanding some basic examples D:

e: [0,1) --> S^1 is not a quotient map. Any neighborhood on the unit circle starting at 1 and going around to e^(i*2pi*c) l be a not open neighborhood (it's complement is not closed) of the unit circle who's preimage is [0,c), which is open in [0,1).

Cool! I believe that this map is not a quotient map. My book goes on to say that:
e: [0,1] -->S^1 is a quotient map because it is also a closed map. Cool, that makes sense to me! I mean before we had the closed neighborhood [a,1) that would map to the not closed neighborhood [e^(i*2pi*a), 1), but now we don't have that problem!

However, to me it still seems that the neighborhood in the unit circle [1,e^(i*2pi*c)) will be a not closed map whose preimage will be [0,c) U {1} which is $NOT$ open either...

Okay so I made the word NOT all fancy because I realized as I was writing this that it was not open because I'm now including the singleton {1}, but I'm going to post this anyway so ya'll can look at my thoughts and give me some feedback as to if my thinking is correct or what not :D

 

[Infrared]

Your observation that the pre image of the arc $\{e^{2\pi i t}:0\leq t<c\}$ for fixed $0<c<1$ is not closed is correct. But this arc is not closed (or open), so we don't have any reason to expect that its pre image should be closed (or open).

 

[PsychonautQQ]

Your observation that the pre image of the arc $\{e^{2\pi i t}:0\leq t<c\}$ for fixed $0<c<1$ is not closed is correct. But this arc is not closed (or open), so we don't have any reason to expect that its pre image should be closed (or open).

Ah shoot. Can you explain to met again why [0,1)---> S^1 is not a quotient map but [0,1] ----> S^1 is?

 

[PsychonautQQ]

Your observation that the pre image of the arc $\{e^{2\pi i t}:0\leq t<c\}$ for fixed $0<c<1$ is not closed is correct. But this arc is not closed (or open), so we don't have any reason to expect that its pre image should be closed (or open).

I think I meant that that in the inverse map from S^1 --> [0,1), the set [0,c) is open but it's pre-image is not.