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$Sp4.1459250_at_phobos.telenet-ops.be>
> (a=1,b=2),
>
> Given
>
> Q * D = V
>
> Q is called locally invertible if there doesn't exist D' != D, such that
>
> Q * D' = V
Date: Mon, 15 Sep 2003 22:15:37 GMT
Message-ID: <dSq9b.22183$Sp4.1459250_at_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