Re: Corrected definition of locally invertible transformation

From: Jan Hidders <jan.hidders_at_pandora.be>
Date: Mon, 15 Sep 2003 22:15:37 GMT
Message-ID: <dSq9b.22183$Sp4.1459250@phobos.telenet-ops.be>

Mikito Harakiri wrote:
> "Jan Hidders" <jan.hidders_at_pandora.be> wrote in message
> news:n9p9b.22067$Wk4.1442847_at_phobos.telenet-ops.be...

>> It is a constant function and therefore independent of everything. The
>> way you formulated your definition all that I have to show is that if you
>> give me a Q and X such that Q(X) is defined then I can construct a
>> transformation Q^-1 that maps Q(X) to X. So if you give me X = {

> (a=1,b=2),

>> (a=1,b=3) } I give you the following transformation:
>>
>>   (SELECT 1 AS a, 2 AS b) UNION (SELECT 1 AS a, 3 AS b)
>>
>> As you can see there is no reference to an instance, and not even to the
>> view. Since I can do this for any X this demonstrates that for any Q and
>> X it holds that Q is locally invertable for X.

>
> Given
>
> Q * D = V
>
> Q is called locally invertible if there doesn't exist D' != D, such that
>
> Q * D' = V

Right! Just to be absolutely sure let me verify the folllowing:

  "globally invertable" = "injective"

Right?

• Jan Hidders

Received on Mon Sep 15 2003 - 17:15:37 CDT