1
$\begingroup$

So I thought as follows: There is a continuous bijection from $[0,1)$ onto $S^1$ for example parametrization $r(x)=(\cos(2\pi x), \sin(2\pi x))$ so if bijective continuous $f:(0,1)\to S^1$ exits then there must be also some $h:(0,1)\to [0,1)$ such that $r \circ h = f$ so it is sufficient to prove that $h$ does not exists, what is easy.

For the second one $[0,1]$ is compact and $S^1$ is Hausdorff as $S^1\subset \Bbb{R}^2$ so if there exists a continuous bijection $g:[0,1]\to S^1$ then it must be homeomorphism, this cannot happened because $[0,1]$ is not homeomorphic to $S^1$

$\endgroup$
  • $\begingroup$ Your argument works fine, indeed it's the standard argument used to prove such results. It remains to fill some details... $\endgroup$ – Crostul Jun 25 '18 at 11:01
  • $\begingroup$ Note that a continuous bijection is not a homeomorphism; the existence of a continuous bijection $A \to B$ does not guarantee the existence of a continuous bijection $B \to A$. $\endgroup$ – Mees de Vries Jun 25 '18 at 11:02
  • $\begingroup$ Why would this $h$ exist??? $\endgroup$ – tomasz Jun 25 '18 at 11:11
  • $\begingroup$ @tomasz that's my question to the community over here. Question is "if my reasoning is all right?" in particular "is the idea that such an h has to exists is right". $\endgroup$ – Kran Jun 25 '18 at 12:29
  • $\begingroup$ @Kacper: this is all hypothethical, so I cannot say "no" -- if you assume falsehood, then everything can be proved. What we can judge is whether or not your reasoning for the existence of $h$ is sound. So what is your reasoning? $\endgroup$ – tomasz Jun 25 '18 at 14:04
0
$\begingroup$

Your argument for the second case is completely correct. While the first case has a bug: even if you use the word "must", the existence of $h$ does not follow immediately from the hypothesis.

A proof of the first case can be done by using the Theorem of invariance of domain which asserts that a continuous inijective map from $\mathbb R^n$ to itself is injective.

Since $S^1$ is locally homeomorphic to $\mathbb R$, then any continuous bijection $f:(0,1)\to S^1$ is open, hence $f^{-1}$ is continuous and $f$ is a homeomorphism. But $S^1$ is not homeo to $(0,1)$ (the former is compact, the later no).

The proof of the invariance of domain in dimension 1 is particularly easy: let $f:(0,1)\to \mathbb R$ continuous an injective. For any $x\in (0,1)$, we want to prove that $f(x)$ is in the interior of the image of $f$. Since $(0,1)$ is connected and $f$ is continuous then $f(0,1)$ is connected, hence an interval I of $\mathbb R$. If $f(x)$ is not an interior point then $I$ is not an open interval, and w.l.o.g. we can suppose $I=[a,b)$ and $f(x)=a$. But now we easily contradict the injectivity of $f$ near $x$ by using that $[x,x+\varepsilon)$ and $(x-\varepsilon,x]$ both maps to intervals (because $f$ is continuous) of the type $[a,something)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.