If the category of elements of a presheave has terminal object, then it is representable.
$begingroup$
Let $F$ be a presheave on a category $C$. If $F$ is representable, then there is a $B in C$ such that $F=Hom(-,B)$. Then, it is not hard to prove that its category of elements $E(F)$ is equivalent to the slice category $C_{/ B}$.
Is it true that if the category of elements $E(F)$ of $F$ has a terminal object, then $F$ is representable? If so, how do I prove it?
category-theory
asked Nov 29 '18 at 15:58
H RH R
1608
$endgroup$
add a comment |
$begingroup$
Let $F$ be a presheave on a category $C$. If $F$ is representable, then there is a $B in C$ such that $F=Hom(-,B)$. Then, it is not hard to prove that its category of elements $E(F)$ is equivalent to the slice category $C_{/ B}$.
Is it true that if the category of elements $E(F)$ of $F$ has a terminal object, then $F$ is representable? If so, how do I prove it?
category-theory
asked Nov 29 '18 at 15:58
H RH R
1608
$endgroup$
add a comment |
$begingroup$
Let $F$ be a presheave on a category $C$. If $F$ is representable, then there is a $B in C$ such that $F=Hom(-,B)$. Then, it is not hard to prove that its category of elements $E(F)$ is equivalent to the slice category $C_{/ B}$.
Is it true that if the category of elements $E(F)$ of $F$ has a terminal object, then $F$ is representable? If so, how do I prove it?
category-theory
asked Nov 29 '18 at 15:58
H RH R
1608
$endgroup$
Let $F$ be a presheave on a category $C$. If $F$ is representable, then there is a $B in C$ such that $F=Hom(-,B)$. Then, it is not hard to prove that its category of elements $E(F)$ is equivalent to the slice category $C_{/ B}$.
Is it true that if the category of elements $E(F)$ of $F$ has a terminal object, then $F$ is representable? If so, how do I prove it?
category-theory
category-theory
asked Nov 29 '18 at 15:58
H RH R
1608
asked Nov 29 '18 at 15:58
H RH R
1608
asked Nov 29 '18 at 15:58
H RH R
1608
asked Nov 29 '18 at 15:58
H RH R
1608
asked Nov 29 '18 at 15:58
H RH R
1608
1608
add a comment |
add a comment |
2 Answers
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$begingroup$
Yes, this is true.
Suppose that $E(F)$ has a terminal object $(B, y)$, and define $theta : F to mathsf{y}(B)$ as follows. Given $A in mathrm{ob}(mathcal{C})$, define
$$theta_A : FA to mathrm{Hom}_{mathcal{C}}(A,B)$$
by letting $theta_A(x)$ be the unique $f : A to B$ given by the (unique) morphism $(A,x) to (B,y)$ in $E(F)$.
Verify that $theta$ is a natural isomorphism.
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
$endgroup$
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
add a comment |
$begingroup$
For a presheaf $F$ on $mathcal C$, recall that $E(F)$ is defined as the category whose objects are $(c,x)$ where $cin mathcal C$ and $xin F(c)$, and whose morphisms $(c,x) to (c',x')$ are those $f:cto c'$ in $mathcal C$ such that $F(f)(x') = x$.
One consequence of the Yoneda lemma is that $$F simeq mathrm{colim}_{(x,c)in E(F)} mathsf y(c) tag{1}$$ (where I use the same notation as Clive for the Yoneda embedding). Now suppose that there is a terminal object $(c_t,x_t)$ in $E(F)$: a colimit taken over a category with a terminal object is just the value of the diagram at that terminal object. Hence $Fsimeq mathsf y (c_t)$.
(Of course the detail of the proof of (1) should fall back to Clive's answer.)
answered Nov 29 '18 at 18:34
PecePece
8,18511241
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2 Answers
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2 Answers
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$begingroup$
Yes, this is true.
Suppose that $E(F)$ has a terminal object $(B, y)$, and define $theta : F to mathsf{y}(B)$ as follows. Given $A in mathrm{ob}(mathcal{C})$, define
$$theta_A : FA to mathrm{Hom}_{mathcal{C}}(A,B)$$
by letting $theta_A(x)$ be the unique $f : A to B$ given by the (unique) morphism $(A,x) to (B,y)$ in $E(F)$.
Verify that $theta$ is a natural isomorphism.
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
$endgroup$
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
add a comment |
$begingroup$
Yes, this is true.
Suppose that $E(F)$ has a terminal object $(B, y)$, and define $theta : F to mathsf{y}(B)$ as follows. Given $A in mathrm{ob}(mathcal{C})$, define
$$theta_A : FA to mathrm{Hom}_{mathcal{C}}(A,B)$$
by letting $theta_A(x)$ be the unique $f : A to B$ given by the (unique) morphism $(A,x) to (B,y)$ in $E(F)$.
Verify that $theta$ is a natural isomorphism.
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
$endgroup$
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
add a comment |
$begingroup$
Yes, this is true.
Suppose that $E(F)$ has a terminal object $(B, y)$, and define $theta : F to mathsf{y}(B)$ as follows. Given $A in mathrm{ob}(mathcal{C})$, define
$$theta_A : FA to mathrm{Hom}_{mathcal{C}}(A,B)$$
by letting $theta_A(x)$ be the unique $f : A to B$ given by the (unique) morphism $(A,x) to (B,y)$ in $E(F)$.
Verify that $theta$ is a natural isomorphism.
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
$endgroup$
Yes, this is true.
Suppose that $E(F)$ has a terminal object $(B, y)$, and define $theta : F to mathsf{y}(B)$ as follows. Given $A in mathrm{ob}(mathcal{C})$, define
$$theta_A : FA to mathrm{Hom}_{mathcal{C}}(A,B)$$
by letting $theta_A(x)$ be the unique $f : A to B$ given by the (unique) morphism $(A,x) to (B,y)$ in $E(F)$.
Verify that $theta$ is a natural isomorphism.
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
edited Nov 29 '18 at 17:00
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
answered Nov 29 '18 at 16:05
Clive NewsteadClive Newstead
50.9k474133
50.9k474133
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
add a comment |
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Could you tell me the definition of $y(B)$?
$endgroup$
– H R
Nov 29 '18 at 16:18
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
$begingroup$
Yes, $mathsf{y}$ is the Yoneda embedding, i.e. $mathsf{y}(B) = mathrm{Hom}_{mathcal{C}}({-}, B)$.
$endgroup$
– Clive Newstead
Nov 29 '18 at 17:00
add a comment |
$begingroup$
For a presheaf $F$ on $mathcal C$, recall that $E(F)$ is defined as the category whose objects are $(c,x)$ where $cin mathcal C$ and $xin F(c)$, and whose morphisms $(c,x) to (c',x')$ are those $f:cto c'$ in $mathcal C$ such that $F(f)(x') = x$.
One consequence of the Yoneda lemma is that $$F simeq mathrm{colim}_{(x,c)in E(F)} mathsf y(c) tag{1}$$ (where I use the same notation as Clive for the Yoneda embedding). Now suppose that there is a terminal object $(c_t,x_t)$ in $E(F)$: a colimit taken over a category with a terminal object is just the value of the diagram at that terminal object. Hence $Fsimeq mathsf y (c_t)$.
(Of course the detail of the proof of (1) should fall back to Clive's answer.)
answered Nov 29 '18 at 18:34
PecePece
8,18511241
$endgroup$
add a comment |
$begingroup$
For a presheaf $F$ on $mathcal C$, recall that $E(F)$ is defined as the category whose objects are $(c,x)$ where $cin mathcal C$ and $xin F(c)$, and whose morphisms $(c,x) to (c',x')$ are those $f:cto c'$ in $mathcal C$ such that $F(f)(x') = x$.
One consequence of the Yoneda lemma is that $$F simeq mathrm{colim}_{(x,c)in E(F)} mathsf y(c) tag{1}$$ (where I use the same notation as Clive for the Yoneda embedding). Now suppose that there is a terminal object $(c_t,x_t)$ in $E(F)$: a colimit taken over a category with a terminal object is just the value of the diagram at that terminal object. Hence $Fsimeq mathsf y (c_t)$.
(Of course the detail of the proof of (1) should fall back to Clive's answer.)
answered Nov 29 '18 at 18:34
PecePece
8,18511241
$endgroup$
add a comment |
$begingroup$
For a presheaf $F$ on $mathcal C$, recall that $E(F)$ is defined as the category whose objects are $(c,x)$ where $cin mathcal C$ and $xin F(c)$, and whose morphisms $(c,x) to (c',x')$ are those $f:cto c'$ in $mathcal C$ such that $F(f)(x') = x$.
One consequence of the Yoneda lemma is that $$F simeq mathrm{colim}_{(x,c)in E(F)} mathsf y(c) tag{1}$$ (where I use the same notation as Clive for the Yoneda embedding). Now suppose that there is a terminal object $(c_t,x_t)$ in $E(F)$: a colimit taken over a category with a terminal object is just the value of the diagram at that terminal object. Hence $Fsimeq mathsf y (c_t)$.
(Of course the detail of the proof of (1) should fall back to Clive's answer.)
answered Nov 29 '18 at 18:34
PecePece
8,18511241
$endgroup$
For a presheaf $F$ on $mathcal C$, recall that $E(F)$ is defined as the category whose objects are $(c,x)$ where $cin mathcal C$ and $xin F(c)$, and whose morphisms $(c,x) to (c',x')$ are those $f:cto c'$ in $mathcal C$ such that $F(f)(x') = x$.
One consequence of the Yoneda lemma is that $$F simeq mathrm{colim}_{(x,c)in E(F)} mathsf y(c) tag{1}$$ (where I use the same notation as Clive for the Yoneda embedding). Now suppose that there is a terminal object $(c_t,x_t)$ in $E(F)$: a colimit taken over a category with a terminal object is just the value of the diagram at that terminal object. Hence $Fsimeq mathsf y (c_t)$.
(Of course the detail of the proof of (1) should fall back to Clive's answer.)
answered Nov 29 '18 at 18:34
PecePece
8,18511241
answered Nov 29 '18 at 18:34
PecePece
8,18511241
answered Nov 29 '18 at 18:34
PecePece
8,18511241
answered Nov 29 '18 at 18:34
PecePece
8,18511241
8,18511241
add a comment |
add a comment |
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