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Definition

$f:X \rightarrow Y$ is called a homeomorphism if :

(a) $f$ is continuous on $X$

(b) $f$ is one to one and onto

(c) $f^{-1} : Y \rightarrow X$ is continuous on $Y$

Then how to construct homeomorphisms $f$ :

(a) $f:[a,b] \rightarrow [c,d]$ where $[a,b],[c,d]$ are closed intervals of $E^1$

(b) $f:(a,b) \rightarrow E^1$

(c) $f:S^3-N \rightarrow E^3$ where $N=(1,0,0,0)$ is the north pole of $S^3$ ( $S$ means sphere)

My attempt :

(a) Consider $X=[-1,1]$ and $Y=[0,5]$

let $f:X \rightarrow Y$ be $f(x)= \frac {5}{2} (x+1)$.

Observe that $f$ is bijective and continuous, being the compositions of addition and multipliction. Moreover, $f^{-1}$ exists and is continuous :

$f^{-1}(x)=\frac {2}{5}x -1$

Note that neiher $[0,1]$ nor $[0,1)$ is homeomorphic as such mapping between these intervals, if constructed, will fail to a bijection due endpoints.

(b) Let $X=(-1,1)$ and $Y=E^1$.

The general open set $(a,b)$ is homeomorphic to $(-1,1)$.

Now define a continuous map $f:(-1,1) \rightarrow E^1$ by

$f(x)=tan \frac{\pi x}{2}$

This continuous bijection possesses a continuous inverse $f^{-1}:E^1 \rightarrow (-1,1)$ by

$f^{-1}(x)= \frac {2}{\pi}arc tan (x)$

Hence $f:(-1,1) \rightarrow E^1$ is a homeomorphism

I changed my question by adding an attempt to my question.

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(a) think linear functions.

(b) think about stuff like $\frac{1}{x}$ or $\tan(x)$

(c) Google "stereographic projection"

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  • $\begingroup$ can you explain or give an example for question (a) ? $\endgroup$ – user326307 May 14 '17 at 4:52
  • $\begingroup$ @user326307 write down the formula for the unique linear function that sends $a$ to $c$ and $b$ to $d$. $\endgroup$ – Henno Brandsma May 14 '17 at 4:54
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – The Count May 14 '17 at 17:39
  • $\begingroup$ @TheCount I gave three (correct) hints. I'm not spelling out the complete answer all the time. $\endgroup$ – Henno Brandsma May 14 '17 at 17:51
  • $\begingroup$ This came up on my review queue and I reviewed it as I felt was best. You may disagree and that's okay. $\endgroup$ – The Count May 14 '17 at 17:52

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