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Finding homeomorphism between topological spaces

  • #1

Homework Statement


show that the two topological spaces are homeomorphic.
71197c5d6992276d973a4fd5ade7fd30.png

Homework Equations


Two spaces are homeomorphic if there exists a continuous bijection with a continuous inverse between them

The Attempt at a Solution


I have tried proving that these two spaces are homeomorphic, however I have no idea whether or not I have done it correctly.
I would really appreciate it if somebody could check my work and give me some tips / comment what I could have done better or different
So I began by drawing both spaces

WP_20171118_23_24_56_Rich.jpg

Then I tried to make a "plan" on how I how transform ##x\rightarrow y##
1. rotate ##X## by ##\frac{\pi}{2}##
2. flatten the bottom part of the circle
3. shrink ##X## so that the top of ##X## touches ##Y##
4. finally shrink ##X## to ##y##

1.
##f(r,\theta)=(r,\theta-\frac{\pi}{2})## this function is continuous and its inverse ##g(r,\theta)=(r,\theta+\frac{\pi}{2})## is also continuous

2.
##f(x,y)=(x,y+\sqrt{1-x^2})## is continuous and its inverse is continuous ##g(x,y)=(x,y-\sqrt{1-x^2})##

3.
##f(x,y)=(x,y/2)## is continuous and its inverse is continuous ##g(x,y)=(x,2y)##

4.
##f(x,y)=(x,\frac{y(-x^4+1)}{\sqrt{1-x^2}})## is continuous and its inverse is continuous ##g(x,y)=(x,\frac{y\sqrt{1-x^2}}{(-x^4+1)})##

as you can probably see I did in 4 steps what my professor would do in 1 maybe 2 functions at most I just find it really hard if I would have tried looking for the function directly.

Also could somebody tell me if this statement is true
the function ##f:\mathbb{R}\rightarrow\mathbb{R}## given with ##f(x)=\arctan{x}## is closed.
Thanks in advance
 

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Answers and Replies

  • #2
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To help other readers:
$$X= \{\,(x,y)\,\vert \,x^2+y^2 < 1\,\} \cup \{\,(x,y)\,\vert \,(x^2+y^2 = 1)\,\wedge \,(x < 0)\,\}\, , \,Y=\{\,(x,y)\,\vert \,0<y\leq -x^4+ 1\,\}$$
I haven't checked your answer, because you didn't mention the spaces you map, which is important for your continuity claims. Also your coordinate change should be a bit more detailed, to guarantee bijectivity which might get lost. Of course one "sees" that both spaces are homeomorphic, but to actually name the bijections can sometimes be a bit complicated. I would stretch ##Y## first and rotate it at last, but the deformation is the crucial point and a bit difficult as you can't just stretch the open boundary to an open half disc. So probably a) stretch ##Y## to a half disc on the left, b) make ##Y## a disc, and c) rotate the thing is the natural way to solve it.

To your last question: Is ##\operatorname{arc tan}## bijective on the branch you define it and is it homeomorphic to the real line?
 
  • #3
Sorry my mistake I probably should have written the spaces I am mapping to.


1.
This function should map from ##X\rightarrow B=\{(x,y)\in\mathbb{R^2}|x^2+y^2 < 1\,\} \cup \{\,(x,y)\,\vert \,(x^2+y^2 = 1)\,\wedge \,(y> 0)\,\}##

##f(r,\theta)=(r,\theta-\frac{\pi}{2})## this function is continuous and its inverse ##g(r,\theta)=(r,\theta+\frac{\pi}{2})## is also continuous

2.
this function maps from ##B\rightarrow C=\{(x,y)\in\mathbb{R^2}|0<y\leq 2\sqrt{1-x^2}\}##
##f(x,y)=(x,y+\sqrt{1-x^2})## is continuous and its inverse is continuous ##g(x,y)=(x,y-\sqrt{1-x^2})##

3.
this function maps from ##C\rightarrow D=\{(x,y)\in\mathbb{R^2}|0<y\leq\sqrt{1-x^2}\}##
##f(x,y)=(x,y/2)## is continuous and its inverse is continuous ##g(x,y)=(x,2y)##

4.
this functions maps from ##D\rightarrow Y##
##f(x,y)=(x,\frac{y(-x^4+1)}{\sqrt{1-x^2}})## is continuous and its inverse is continuous ##g(x,y)=(x,\frac{y\sqrt{1-x^2}}{(-x^4+1)})##

For the ##\arctan## I believe the answer is no. it is injective however it is not a surjective function from ##f:\mathbb{R}\rightarrow\mathbb{R}## and if we take ##\mathbb{R}## which is a closed set it is mapped to ##(\frac{-\pi}{2},\frac{\pi}{2})## which is an open set in ##\mathbb{R}##
 
  • #4
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This looks correct, although formally you would have to show that the codomains are identical to the stated spaces, but this would simply be - as a little remark, why ##x=1## isn't in ##D## and ##f_4## is thus continuous - a nice service for your readers. You could also compose all functions and their inverses to get the one homeomorphism you were looking for. If nothing else it'll be a funny monstrous formula.

Of course ##\operatorname{arc tan}## isn't surjective on ##\mathbb{R}##, but this isn't the crucial point here. Usually we look at ##f\, : \,X \twoheadrightarrow f(X)## to investigate a function, because the trivial case ##f(X) \subsetneq Y## doesn't say much about ##f##. So basically, you are right: not surjective. But what happens if you consider ##\operatorname{arc tan}\, : \,\mathbb{R} \rightarrow (-\dfrac{\pi}{2},\dfrac{\pi}{2})## instead? Is this interval as topological space homeomorph to ##\mathbb{R}\,##? The graph of ##\operatorname{arc tan}## and the real line are topologically the same thing, so why shouldn't it be a closed function?
 
  • #5
You are correct. Those two spaces ##\mathbb{R}## and ##(-\frac{\pi}{2},\frac{\pi}{2})## are indeed homeomorphic and ##\tan^{-1}## is a continuous bijection. ##\tan^{-1}## is also closed and open since it gives a homeomorphism between ##\mathbb{R}## and ##(-\frac{\pi}{2},\frac{\pi}{2})##.
However there is still one thing that is bothering me. So whenever I am given a function ##f:\mathbb{R}\rightarrow\mathbb{R}## and asked to check whether it is open or closed I should only look at its codomain? It was only bothering me because I started thinking that since ##\mathbb{R}## is a closed set that its map ##(-\frac{\pi}{2},\frac{\pi}{2})## should also be a closed set under##\mathbb{R}##
 
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  • #6
Alternately could I have just checked if random sequence that converges to x its f- values converge to ##f(x)## which must be in the codomain and if that be the case from all convergent sequences the map would be closed?
 
  • #7
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You are correct. Those two spaces ##\mathbb{R}## and ##(-\frac{\pi}{2},\frac{\pi}{2})## are indeed homeomorphic and ##\tan^{-1}## is a continuous bijection. ##\tan^{-1}## is also closed and open since it gives a homeomorphism between ##\mathbb{B}## and ##(-\frac{\pi}{2},\frac{\pi}{2})##.
However there is still one thing that is bothering me. So whenever I am given a function ##f:\mathbb{R}\rightarrow\mathbb{R}## and asked to check whether it is open or closed I should only look at its codomain?
No, because you were right. Your question means basically: Is a function ##f\, : \,X \rightarrow Y## open (closed), if it is open (closed) restricted to its codomain? From left to right this is true, but not the other way around. Say ##X = \mathbb{R}^{>0}\; , \;Y=\mathbb{R}## and ##f(x)=\dfrac{x}{\,\vert \,x\,\vert \,}\,.## Then ##f(X)=(0,1]## is trivially closed in the codomain ##f(X)## but not in ##Y##. And the same is true for ##\tan^{-1}##. The image of the closed interval ##(-\infty,0]## is ##(-\dfrac{\pi}{2},0]## which is closed in the codomain ##(-\dfrac{\pi}{2},\dfrac{\pi}{2})## and not closed in ##\mathbb{R}##.
It was only bothering me because I started thinking that since ##\mathbb{R}## is a closed set that its map ##(-\frac{\pi}{2},\frac{\pi}{2})## should also be a closed set under##\mathbb{R}##
Correct, this is of course not closed in ##\mathbb{R}##.

A good example how the topological space considered makes a huge difference: ##\mathbb{R} \cong \tan^{-1}(\mathbb{R}) = (-\frac{\pi}{2},\frac{\pi}{2})## and ##\mathbb{R} \ncong_{\operatorname{arc tan}} \mathbb{R}##, so closed on the induced topology of the codomain but not closed on the entire space.
... I shouldn't answer topological question, there are far too many pitfalls and one has to be extremely cautious as intuition isn't of much value.
 
  • #8
I though I started understanding this but then I saw this post in the Wikipedia site which gave an example of a closed map
d202f9b8d9f459cc97246e4d1a703ef8.png

So I guess we should look on the induced topology afterall
 

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  • #9
Just to be sure I'm going to ask one of the professors tomorrow just to see what they believe to be the correct answer
 
  • #10
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I though I started understanding this but then I saw this post in the Wikipedia site which gave an example of a closed map View attachment 215272
So I guess we should look on the induced topology after
##f\, : \,\mathbb{R} \longrightarrow \mathbb{R}## with ##f\, : \, x \longmapsto x^2## is continuous and closed intervals map onto closed intervals, but ##f((-1,1))=[0,1)## which is not open.
 
  • #11
I know that however could you explain why it is closed? since ##f(\mathbb{R})=[0,\infty)##and isn't this not closed in ##\mathbb{R}## ?
 
  • #12
Note to self ##[0,\infty)## is indeed closed
 

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