What is the relationship between external and internal composition in a cartesian closed category?Is there a...
What is the relationship between external and internal composition in a cartesian closed category?
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$begingroup$
Denote "the" category of sets and functions by $S$. The hom set of functions from set $X$ to set $Y$ is denoted by $S(X,Y)$.
If $C$ is a cartesian closed category denote by $C(x,y)$ the set of morphisms from $x$ to $y$ in $C$. In such a $C$ there exists a natural bijection between $C(x,y)$ and $C(1,y^x)$. In a sense, $y^x$ reifies inside $C$ the set $C(x,y)$ in $S$. Both $S(1,C(x,y))$ and $C(1,y^x)$ are sets, and in particular, if $x=1$, then both $S(1,C(1,y))$ and $C(1,y^1)$ are sets.
Anyhow, how does the "external" law of composition $C(x,y) times C(y,z) to C(x,z)$ in $S$ of $C$ relate to the "internal" law of composition $y^x times z^y to z^x$ in $C$? In summary, does the internal composition "reify" the external composition?
ct.category-theory
edited 6 hours ago
Alex Kruckman
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asked 9 hours ago
Ellis D CooperEllis D Cooper
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$endgroup$
add a comment |
$begingroup$
Denote "the" category of sets and functions by $S$. The hom set of functions from set $X$ to set $Y$ is denoted by $S(X,Y)$.
If $C$ is a cartesian closed category denote by $C(x,y)$ the set of morphisms from $x$ to $y$ in $C$. In such a $C$ there exists a natural bijection between $C(x,y)$ and $C(1,y^x)$. In a sense, $y^x$ reifies inside $C$ the set $C(x,y)$ in $S$. Both $S(1,C(x,y))$ and $C(1,y^x)$ are sets, and in particular, if $x=1$, then both $S(1,C(1,y))$ and $C(1,y^1)$ are sets.
Anyhow, how does the "external" law of composition $C(x,y) times C(y,z) to C(x,z)$ in $S$ of $C$ relate to the "internal" law of composition $y^x times z^y to z^x$ in $C$? In summary, does the internal composition "reify" the external composition?
ct.category-theory
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
$endgroup$
$begingroup$
I've edited to add MathJax markup to your question.
$endgroup$
– Alex Kruckman
7 hours ago
add a comment |
$begingroup$
Denote "the" category of sets and functions by $S$. The hom set of functions from set $X$ to set $Y$ is denoted by $S(X,Y)$.
If $C$ is a cartesian closed category denote by $C(x,y)$ the set of morphisms from $x$ to $y$ in $C$. In such a $C$ there exists a natural bijection between $C(x,y)$ and $C(1,y^x)$. In a sense, $y^x$ reifies inside $C$ the set $C(x,y)$ in $S$. Both $S(1,C(x,y))$ and $C(1,y^x)$ are sets, and in particular, if $x=1$, then both $S(1,C(1,y))$ and $C(1,y^1)$ are sets.
Anyhow, how does the "external" law of composition $C(x,y) times C(y,z) to C(x,z)$ in $S$ of $C$ relate to the "internal" law of composition $y^x times z^y to z^x$ in $C$? In summary, does the internal composition "reify" the external composition?
ct.category-theory
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
$endgroup$
Denote "the" category of sets and functions by $S$. The hom set of functions from set $X$ to set $Y$ is denoted by $S(X,Y)$.
If $C$ is a cartesian closed category denote by $C(x,y)$ the set of morphisms from $x$ to $y$ in $C$. In such a $C$ there exists a natural bijection between $C(x,y)$ and $C(1,y^x)$. In a sense, $y^x$ reifies inside $C$ the set $C(x,y)$ in $S$. Both $S(1,C(x,y))$ and $C(1,y^x)$ are sets, and in particular, if $x=1$, then both $S(1,C(1,y))$ and $C(1,y^1)$ are sets.
Anyhow, how does the "external" law of composition $C(x,y) times C(y,z) to C(x,z)$ in $S$ of $C$ relate to the "internal" law of composition $y^x times z^y to z^x$ in $C$? In summary, does the internal composition "reify" the external composition?
ct.category-theory
ct.category-theory
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
edited 6 hours ago
Alex Kruckman
1,9791 gold badge12 silver badges16 bronze badges
1,9791 gold badge12 silver badges16 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
asked 9 hours ago
Ellis D CooperEllis D Cooper
163 bronze badges
163 bronze badges
$begingroup$
I've edited to add MathJax markup to your question.
$endgroup$
– Alex Kruckman
7 hours ago
add a comment |
$begingroup$
I've edited to add MathJax markup to your question.
$endgroup$
– Alex Kruckman
7 hours ago
$begingroup$
I've edited to add MathJax markup to your question.
$endgroup$
– Alex Kruckman
7 hours ago
$begingroup$
I've edited to add MathJax markup to your question.
$endgroup$
– Alex Kruckman
7 hours ago
add a comment |
1 Answer
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$begingroup$
The internal composition $y^xtimes z^yto z^x$ induces, by composition, $C(1,y^xtimes z^y)to C(1,z^x)$. The domain of this morphism is naturally equivalent to $C(1,y^x)times C(1,z^y)$, by definition of product in $C$. So we get a function $C(1,y^x)times C(1,z^y)to C(1,z^x)$, which, as you noted, is equivalent to $C(x,y)times C(y,z)to C(x,z)$. And that's the external composition.
answered 8 hours ago
Andreas BlassAndreas Blass
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$endgroup$
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$begingroup$
The internal composition $y^xtimes z^yto z^x$ induces, by composition, $C(1,y^xtimes z^y)to C(1,z^x)$. The domain of this morphism is naturally equivalent to $C(1,y^x)times C(1,z^y)$, by definition of product in $C$. So we get a function $C(1,y^x)times C(1,z^y)to C(1,z^x)$, which, as you noted, is equivalent to $C(x,y)times C(y,z)to C(x,z)$. And that's the external composition.
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
$endgroup$
add a comment |
$begingroup$
The internal composition $y^xtimes z^yto z^x$ induces, by composition, $C(1,y^xtimes z^y)to C(1,z^x)$. The domain of this morphism is naturally equivalent to $C(1,y^x)times C(1,z^y)$, by definition of product in $C$. So we get a function $C(1,y^x)times C(1,z^y)to C(1,z^x)$, which, as you noted, is equivalent to $C(x,y)times C(y,z)to C(x,z)$. And that's the external composition.
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
$endgroup$
add a comment |
$begingroup$
The internal composition $y^xtimes z^yto z^x$ induces, by composition, $C(1,y^xtimes z^y)to C(1,z^x)$. The domain of this morphism is naturally equivalent to $C(1,y^x)times C(1,z^y)$, by definition of product in $C$. So we get a function $C(1,y^x)times C(1,z^y)to C(1,z^x)$, which, as you noted, is equivalent to $C(x,y)times C(y,z)to C(x,z)$. And that's the external composition.
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
$endgroup$
The internal composition $y^xtimes z^yto z^x$ induces, by composition, $C(1,y^xtimes z^y)to C(1,z^x)$. The domain of this morphism is naturally equivalent to $C(1,y^x)times C(1,z^y)$, by definition of product in $C$. So we get a function $C(1,y^x)times C(1,z^y)to C(1,z^x)$, which, as you noted, is equivalent to $C(x,y)times C(y,z)to C(x,z)$. And that's the external composition.
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
answered 8 hours ago
Andreas BlassAndreas Blass
59.4k7 gold badges144 silver badges231 bronze badges
59.4k7 gold badges144 silver badges231 bronze badges
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$begingroup$
I've edited to add MathJax markup to your question.
$endgroup$
– Alex Kruckman
7 hours ago