Re: Is the analytic image of an open set always an open set?

On Thu, 12 May 2005 18:51:34 +0200, Jannick Asmus
<jannick.news@xxxxxx> wrote:

>On 12.05.2005 15:41, David C. Ullrich wrote:
>> On Thu, 12 May 2005 08:46:53 +0200, Jannick Asmus
>> <jannick.news@xxxxxx> wrote:
>>
>>
>>>On 12.05.2005 08:39, Snis Pilbor wrote:
>>>
>>>>Hi,
>>>>
>>>> A friend of mine was very proud to show off a very simple elegant
>>>>proof of the max principal. Unfortunately we realized it was faulty
>>>>because it assumed the continuous image of an open set is open, which
>>>>isn't necessary true (eg abs(R)=[0,oo)). The proof could still work
>>>>though if it is true that the analytic image of an open set is always
>>>>open. Is this the case? We tried to come up with a proof or
>>>>counterexample but could not. How would one prove this? This is not
>>>>homework.
>>>>
>>>> Snis.
>>>>
>>>
>>>This is true if the function is not locally constant.
>>>
>>>For a proof take into account that (1) the property of being an open map
>>>is local and (2) any analytic map can be locally represented by z -> z^n
>>>(for some natural number n>0).
>>
>>
>> I think it's possible that this is going to cause some confusion,
>> unless you clarify what you mean by "represented by".
>>
>> (Of course it's true, with a suitable interpretation of that phrase.
>> But I don't know whether there's a proof of this fact that doesn't
>> already contain a proof of the open mapping theorem, implicitly...)
>
>In order to clarify what you deemed to be possibly misleading, I was
>referring to the fact that any analytic map f can be locally represented
>as f = g1 o f_n o g2 with appropriate biholomorphic mappings g1, g2 and
>f_n(z)=z^n (n e N).
>
>The proof of this claim is certainly independent of the open mapping
>theorem. It uses the expansion theorem and the inverse mapping theorem
>of (real) calculus (yielding a biholomorphic map in our case).

There's no point in fighting about this. But my point was precisely
that if we know the expansion theorem and the real inverse function
theorem then the open mapping theorem is an immediate consequence
(as is the more precise result about f being represented as z^n.)

>After these preliminary remarks, the map of the original problem can be
>locally written as f_n o g with some biholomorphic map g and f_n(z)=z^n
>(n>0). With this, the claim easily follows.
>
>After all, I could think of a proof using Rouché's theorem. It is
>logically based on the residue theorem. Since this is commonly presented
>later in a course, the assertion of this posting can be proved with the
>theorems on the local behaviour of analytic maps. As the open mapping
>theorem covers basically a local property of mappings, I prefer
>considering the claim above to be a corollary of those local properties.

I don't know what's "common", but in my favorite book (Rudin) and
in the book commonly used as the text for the course here (Conway)
the open mapping theorem is indeed proved via the residue theorem,
or rather using the special case that says that the number of
zeroes of a function in a disk is given by a certain integral
on the boundary. There's nothing non-local about that - the
global version of the relevant integral theorems is not needed.

>J.
>
>>
>>
>>>Since the latter maps are open, we're done.
>>>
>>>Cheers,
>>>J.
>>
>>
>>
>> ************************
>>
>> David C. Ullrich


************************

David C. Ullrich
.

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