Changes http://page.axiom-developer.org/zope/mathaction/187TroubleWithTuples/diff --
++added: ??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: $$g(d) = (d,d)$$ ++added: ++added: -- forwarded from http://page.axiom-developer.org/zope/mathaction/[EMAIL PROTECTED] _______________________________________________ Axiom-developer mailing list [email protected] http://lists.nongnu.org/mailman/listinfo/axiom-developer
