Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $U$ and $U'$ be two domains in $\Bbb C$, and $f$ be a homeomorphism in $U$ and $U'$ then domain $U$ is simply connected $\iff$ $U'$ is simply connected. I found this problem in complex analysis. So I would prefer to know its proof from complex point of view rather using topological propositions. Thanks.

There are few properties which are equivalent for a domain $D$ in Complex plane.

a)$D$ is simply connected.

b)for each $z_0\in \Bbb C$\ $D$ there is a analytic branch of $log(z-z_0)$ defined on $D$.

c)The compliment of $D$ in the extended complex plane $\Bbb C^*$ is connected.

share|improve this question
3  
Any simply connected domain in $\mathbb{C}$ is conformally equivalent to the unit disc, by the Riemann mapping theorem. –  jmracek Dec 11 '13 at 22:28
1  
Unless it is $\mathbb{C}$ itself ! If you want to prove this result with complex analysis, you should at least choose your $f$ to be biholomorphic. –  Plop Dec 11 '13 at 22:31
1  
@jmracek Not any, there is an exception. But that exception is no problem here as a homeomorphism to the unit disc is readily available. –  Hagen von Eitzen Dec 11 '13 at 22:31
1  
Whoops! Of course I meant any simply connected domain which is not all of $\mathbb{C}$. Thanks for the correction guys. –  jmracek Dec 11 '13 at 22:32
1  
You should not expect to be able to use complex analysis to establish anything about a function which might be nowhere differentiable. A homeomorphism can be a truly bazaar thing, and not be subject to any "analysis" at all. –  Steven Gubkin Dec 11 '13 at 23:58
show 2 more comments

4 Answers

up vote 2 down vote accepted

What you say is true just by writing down the definition of homeomorphism and the definition of simply connected essentially. Simply connected is a topological property, and homeomorphisms let you transfer topological properties between spaces.

What complex analysis does beautifully is to show that any two simply connected proper subsets of the plane are homeomorphic. This is a really nontrivial result, and the Riemann mapping theorem solves it in a spectacular way: by proving the much stronger result that they are biholomorphic!

share|improve this answer
    
Finally I got your point. Thanks you so much Steven. –  Mat He Mat Cian Dec 12 '13 at 3:27
add comment

We will use the Riemann criterion for complex simply connected spaces.

Let $U$ be simply connected and $f$ be biholomorphic.

We have only to prove that if $g:U'\to \Bbb C$ holomorphic with $g(z)\neq 0$ for every $z$, then there is a holomorphic $g_1:U'\to \Bbb C:g(z)=g_1^2(z)$ for every $z$.

Let such a holomorphic function $g$.The composition $gof:U\to \Bbb C$ is holomorphic and $gof(w)\neq 0$ for every $w$. Because $U$ is simply connected ,there is a holomorphic $h:U\to \Bbb C:gof(w)=h^2(w)$ for every $w\in U$.

Let $hof^{-1}:U'\to \Bbb C$. Then it is holomorphic and for every $z\in U'$ we have that $(hof^{-1}(z))^2=(h(f^{-1}(z))^2=g(z)$ for every $z$.

share|improve this answer
2  
This answer also assumes that $f$ is a biholomorphism, which was not stated in the statement of the problem. –  Steven Gubkin Dec 11 '13 at 23:04
    
you are right. I just noticed he is also using biholomorphism. –  Mat He Mat Cian Dec 11 '13 at 23:32
    
@StevenGubkin Correct. But if you want to prove it with complex analysis only,the only way that i can think of is that $f$ is biholomorphic. –  Mitsos Dec 11 '13 at 23:45
    
@MatHeMatCian Correct. But if you want to prove it with complex analysis only,the only way that i can think of is that $f$ is biholomorphic. –  Mitsos Dec 11 '13 at 23:46
    
Indeed, this is why it is really a flawed question. –  Steven Gubkin Dec 11 '13 at 23:56
add comment

The question is quite weird ; the notion of simply connectedness is a topological notion and the property you are asking is a lot simpler than complex analysis. If $U$ and $U'$ are homeomorphic , then any topological notion on $U$ can be carried to $U'$ by the homeomorphism.

For your question, this can take the following form. Assume that $U$ is simply connected. Choose a point $y$ in $U'$ and take a loop $\gamma(t)$ with base-point $y$ in $U'$. Then $f^{-1} \circ \gamma(t)$ is a loop in $U$ with base-point $x = f^{-1}(y)$. Since $U$ is simply connected, you have an homotopy $F(s,t)$ from $f^{-1} \circ \gamma$ to the constant loop based at $x$. Then $f \circ F(s,t)$ is a homotopy from $\gamma$ to the constant loop based at $y$.

share|improve this answer
add comment

It seems to me like an incredible overkill to use the Riemann mapping theorem here. Perhaps you can use the following theorem:

Let $U$ be a connected open subset of $\mathbf C$. Then $U$ is simply-connected if and only if for every holomorphic function $g$ on $U$ and every closed path $C$ in $U$, $\int_C g(z) dz = 0$.

Then, under the assumption that $U$ is simply connected, use $f$ and the change-of-variables formula to prove that the same criterion holds in $U'$.

(Of course, the real reason is a purely topological one: two homeomorphic spaces have isomorphic fundamental groups. I'm only giving this weird solution because you insist on avoiding "topological" arguments.)

share|improve this answer
2  
You cannot use change of variables formula if $f$ is merely continuous: it may be nowhere differentiable for instance. –  Steven Gubkin Dec 11 '13 at 22:49
    
Hmm that is true. For some reason I read the question as if $f$ were a biholomorphism. –  Bruno Joyal Dec 11 '13 at 22:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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