1

I am new to topology...so the question might be very trivial and stupid.

Is the function $f:\mathbb{R}\to S^1 $(unit circle) such that $x\mapsto (sin(x),cox(x))$ open or closed?

So first I assume that $S^1$ is equipped with the relative topology. Thus any open set in this topology is an intersection of open balls in $\mathbb{R}^2$ with $S^1$.

Then I think...if we take $(0,\pi)$ in $\mathbb{R}$, then the image of $(0,\pi)$ under $f$ should be $(0,1]\times (-1,1)$ in $S^1$, which is not open since $(0,1]$ is not?

But then my confusion is that I actually only know that $(0,1]$ is not open in $\mathbb{R}$, how can I connect this to the relative topology of $S^1$ in $\mathbb{R}^2$?

Did I make some trivial mistakes? Could you please help me? Also to prove if this function is closed or not...I have on clue...

Thanks!

Tortuga
  • 477
  • It is both open and closed. Your map factors through the quotient $\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$ this quotient is open and closed and the other map $\mathbb{R}/\mathbb{Z}\rightarrow S^1$ is an homeomorphism. – Yanko Sep 21 '18 at 18:35
  • @Yanko Thanks for your comment! But I have not learnt anything about the quotient topology...and how can I see that the map you gave is a homeomorphism? – Tortuga Sep 21 '18 at 18:43
  • Let's go slowly. First do you understand why the map $x\mapsto (\cos(x),\sin(x))$ is equivalent to the composition $x\mapsto x+2\pi\mathbb{ Z} \mapsto (cos(x),sin(x))$ ? (The second map is well defined because $(\cos 2\pi n, \sin 2\pi n) = (0,0)$) – Yanko Sep 21 '18 at 18:45
  • 3
    @Yanko It's not closed. Consider the image of the closed set ${2\pi n+1/n:n=1,2,\dots }.$ – zhw. Sep 21 '18 at 18:46
  • @Yanko Yes, I understand, I think this is because that both $cos$ and $sin$ are $2\pi$-periodic – Tortuga Sep 21 '18 at 18:48
  • @Abigail Sorry Abigail but looks like I'm wrong. check out zhw.'s example – Yanko Sep 21 '18 at 18:58
  • @Yanko, thanks! I will think of it, but how about the openness of this function? Am I right or I am still misunderstanding what is going on, thanks – Tortuga Sep 21 '18 at 19:00
  • @Abigail I think your confusion makes sense, in fact I believe that my arguments works (for the open part) so it is open... – Yanko Sep 21 '18 at 19:02
  • @Abigail I found this question https://math.stackexchange.com/questions/1355730/proof-that-bbb-r-bbb-z-is-isomorphic-to-s1 – Yanko Sep 21 '18 at 19:14
  • @Abigail If it convinces you I can easily show that the map is open – Yanko Sep 21 '18 at 19:15
  • @Yanko oops…I was over thinking the map you mentioned. Yes, I agree that we have an isomorphism from $\mathbb{R}/\mathbb{Z}\to S^1$, what should I do next? Thanks – Tortuga Sep 21 '18 at 19:17
  • @Abigail the point is that it's an isomorphism of topological groups (i.e. homeomorphism) and the map $\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}$ is always open (quotient maps are open by definition of the quotient topology) so you have a composition of open maps. – Yanko Sep 21 '18 at 19:19
  • @Yanko Once we can prove the inverse function is still continuous...I will do some research on quotient topology. Thanks! – Tortuga Sep 21 '18 at 19:21
  • @Abigail this question is a VERY general way to show that (continuous) isomorphisms are in fact open, but I believe in your case there must be an easier way I'm working on it https://math.stackexchange.com/questions/287128/continuous-homomorphism-into-locally-compact-hausdorff-group – Yanko Sep 21 '18 at 19:21
  • @Yanko Hi, to prove f is not close, I think the image of the set is ${sin(1/n)}\times {cos(1/n)}$. Could you give me a hint why it is not closed in $S^1$? Is it because the closure of this set should contain $0\times 1$? – Tortuga Sep 21 '18 at 19:34
  • @Abigail That's right. – Yanko Sep 21 '18 at 21:04
  • @Yanko. See my Answer – DanielWainfleet Sep 21 '18 at 21:39
  • 1
    @DanielWainfleet Looks good! – Yanko Sep 21 '18 at 21:40
  • @Yanko I don't think quotient maps are always open (in general) either. For example, the quotient of $[0,1]$ by identifying $1/2$ with $1$ results in a $\rho$ shape, and the image of $(1/4, 3/4)$ will not be an open subset. – Daniel Schepler Sep 21 '18 at 22:27
  • Oh, for that matter, the restriction of $f$ to $[0,1]$ is also a quotient map, but it isn't open because the image of $[0, 1/2)$ in $S^1$ isn't open. – Daniel Schepler Sep 21 '18 at 22:33
  • One useful way to prove a function $f$ is open: Suppose $f : X \to Y$ and $X$ has a given system of neighborhood bases. Then $f$ is open if and only if for every $x_0 \in X$ and basic neighborhood $N$ of $x_0$, we have $f(N)$ is a neighborhood of $f(x_0)$. – Daniel Schepler Sep 21 '18 at 22:39
  • @DanielSchepler You are right, the actual true statement is that any group quotient homomorphism is open. (at least under some assumptions, for example that the group is locally compact) – Yanko Sep 21 '18 at 22:53

1 Answers1

1

An open subset of $\Bbb R$ is a union of open intervals and an open interval is a union of open intervals of length less than $2\pi.$ If $a<b<a+2\pi$ then the image of $(a,b)$ is $\{(\sin (a+x), \cos (a+x)): 0<x<b-a\},$ which is open in $S^1.$

DanielWainfleet
  • 57,985
  • Thanks for your answer! It makes sense. Could you please tell me where is my mistake in OP? I should not conclude $ (0,1]\times(−1,1)$ is open in $S^1$, should I? Thanks! – Tortuga Sep 22 '18 at 04:13
  • @Abigail draw this, it turns out to be open in $S^1$. In fact $S^1\cap (0,1]\times(-1,1) = S^1\backslash {-1,i,-i}$ – Yanko Sep 22 '18 at 13:30
  • @Yanko Thanks a lot! – Tortuga Sep 22 '18 at 16:33