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

[Bill Page]

Changes 
http://page.axiom-developer.org/zope/mathaction/187TroubleWithTuples/diff
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??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:
$$g(d) = (d,d)$$

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