Finding homeomorphism between topological spaces

[nightingale123]

Homework Statement

show that the two topological spaces are homeomorphic.

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

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

 

[fresh_42]

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?

 

[nightingale123]

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 $\mathbb{R}$

 

[fresh_42]

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?

 

[nightingale123]

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$\mathbb{R}$

 

[nightingale123]

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?

 

[fresh_42]

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.

 

[nightingale123]

I though I started understanding this but then I saw this post in the Wikipedia site which gave an example of a closed map

So I guess we should look on the induced topology afterall

 

[nightingale123]

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

 

[fresh_42]

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.

 

[nightingale123]

I know that however could you explain why it is closed? since $f(\mathbb{R})=[0,\infty)$and isn't this not closed in $\mathbb{R}$ ?

 

[nightingale123]

Note to self $[0,\infty)$ is indeed closed