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[Axiom-developer] [#187 trouble with tuples]

 From: Bill Page Subject: [Axiom-developer] [#187 trouble with tuples] Date: Mon, 04 Jul 2005 12:23:00 -0500

Changes
http://page.axiom-developer.org/zope/mathaction/187TroubleWithTuples/diff
--

??changed:
-the same category. If $f:A \times B \rightarrow C$ is a mapping, where $A$, $B$
-are from the same category, we may sometimes let $D = A \times B$ and identify
-$f$ as $f:D \rightarrow C$ (let me rename this to $g:D \rightarrow C$).
However,
-there is this subtle distinction in the way we give the definition of $f$ and
-$g$. In the first case, we would write f(a,b) = c, where as in the second case,
-we would write g(d) = c, with d = (a,b). The two are *not* equivalent as
-*mappings*: $f$ is binary and $g$ is unary. To define $c$ to be $a+b$ in both
-cases, say, it is straight forward in the first case $f(a,b)=a+b$. In the
second
-case, there is necessarily a composition with two projection maps $p:D -\rightarrow A$ and $q:D \rightarrow B$, where $p(d)=a$, $q(d) = b$. The true
-definition of $g$ is: $g(d) = p(d)+q(d)$. If the target $C$ is more involved,
-say $C$ is D^2$and$f$is meant to be the diagonal map$D \rightarrow D^2$, -then the$g$-form would be more preferrable:$g(d) = (d,d)$. the same category. If $$f:A \times B \rightarrow C$$ is a mapping, where$A$,$B$are from the same category, we may sometimes let $$D = A \times B$$ and identify$f$as $$f:D \rightarrow C$$ let me rename this to: $$g:D \rightarrow C$$. However, there is this subtle distinction in the way we give the definition of$f$and$g$. In the first case, we would write f(a,b) = c, where as in the second case, we would write g(d) = c, with d = (a,b). The two are *not* equivalent as *mappings*:$f$is binary and$g$is unary. To define$c$to be$a+b$in both cases, say, it is straight forward in the first case $$f(a,b)=a+b$$ In the second case, there is necessarily a composition with two projection maps $$p:D \rightarrow A$$ and $$q:D \rightarrow B$$ where $$p(d)=a$$ $$q(d) = b$$ The true definition of$g$is: $$g(d) = p(d)+q(d)$$ If the target$C$is more involved, say$C$is D^2$ and $f$ is meant to be the
diagonal map
$$D \rightarrow D^2$$
then the $g$-form would be more preferrable:
$$g(d) = (d,d)$$