There is a natural morphism $limlimits_{longleftarrow} beta to limlimits_{longleftarrow} beta circ...
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This is from Categories & Sheaves by Kashiwara & Schapira.
$$text{Hom}_{text{Set}}(X, lim_{leftarrow} beta) xrightarrow{sim} lim_{leftarrow} text{Hom}_{text{Set}}(X, beta) tag{2.1.1}$$
Here's the exercise:
Let $varphi : J to I$ and $beta : I^{op} to text{Set}$ be
functors. Denote by $varphi^{op} : J^{op} to I^{op}$ the associated
functor. Using (2.1.1), we get a natural morphism:
$$ lim_{longleftarrow} beta to lim_{longleftarrow}beta circvarphi^{op} $$
Firstly, I don't see how the left or right is a functor so how can we speak of the natural map?
Secondly, how do you make the map? If we have a natural map $a in limlimits_{longleftarrow} beta$ such that $beta(f) circ a_i = a_j$ for any $f : j to i$ in $I$, I'm not seeing how this should work.
category-theory limits-colimits functors hom-functor
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
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show 1 more comment
up vote
2
down vote
favorite
This is from Categories & Sheaves by Kashiwara & Schapira.
$$text{Hom}_{text{Set}}(X, lim_{leftarrow} beta) xrightarrow{sim} lim_{leftarrow} text{Hom}_{text{Set}}(X, beta) tag{2.1.1}$$
Here's the exercise:
Let $varphi : J to I$ and $beta : I^{op} to text{Set}$ be
functors. Denote by $varphi^{op} : J^{op} to I^{op}$ the associated
functor. Using (2.1.1), we get a natural morphism:
$$ lim_{longleftarrow} beta to lim_{longleftarrow}beta circvarphi^{op} $$
Firstly, I don't see how the left or right is a functor so how can we speak of the natural map?
Secondly, how do you make the map? If we have a natural map $a in limlimits_{longleftarrow} beta$ such that $beta(f) circ a_i = a_j$ for any $f : j to i$ in $I$, I'm not seeing how this should work.
category-theory limits-colimits functors hom-functor
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
Surely there must be additional conditions on $phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton.
– Arnaud D.
Jun 12 at 14:36
1
Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $Xmapsto lim mathrm{hom}(X,beta)$ as $lim beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit.
– asdq
Jun 12 at 14:51
It's on page 36.
– Roll up and smoke Adjoint
Jun 12 at 14:53
On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $varprojlimbeta$ a functor of $beta$, in other words $varprojlim:operatorname{Fun}(I,C)to C$. Similarly if $C$ has all $J$ limits, you have a functor $operatorname{Fun}(I,C)to C$ such that $betamapstovarprojlimbetacircvarphi^{op}$. The claim is that this is a natural transformation.
– Roland
Jun 12 at 17:12
@Roland I see that part now, thank you. How do I show that there is a map between the two though?
– Roll up and smoke Adjoint
Jun 12 at 17:49
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is from Categories & Sheaves by Kashiwara & Schapira.
$$text{Hom}_{text{Set}}(X, lim_{leftarrow} beta) xrightarrow{sim} lim_{leftarrow} text{Hom}_{text{Set}}(X, beta) tag{2.1.1}$$
Here's the exercise:
Let $varphi : J to I$ and $beta : I^{op} to text{Set}$ be
functors. Denote by $varphi^{op} : J^{op} to I^{op}$ the associated
functor. Using (2.1.1), we get a natural morphism:
$$ lim_{longleftarrow} beta to lim_{longleftarrow}beta circvarphi^{op} $$
Firstly, I don't see how the left or right is a functor so how can we speak of the natural map?
Secondly, how do you make the map? If we have a natural map $a in limlimits_{longleftarrow} beta$ such that $beta(f) circ a_i = a_j$ for any $f : j to i$ in $I$, I'm not seeing how this should work.
category-theory limits-colimits functors hom-functor
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
This is from Categories & Sheaves by Kashiwara & Schapira.
$$text{Hom}_{text{Set}}(X, lim_{leftarrow} beta) xrightarrow{sim} lim_{leftarrow} text{Hom}_{text{Set}}(X, beta) tag{2.1.1}$$
Here's the exercise:
Let $varphi : J to I$ and $beta : I^{op} to text{Set}$ be
functors. Denote by $varphi^{op} : J^{op} to I^{op}$ the associated
functor. Using (2.1.1), we get a natural morphism:
$$ lim_{longleftarrow} beta to lim_{longleftarrow}beta circvarphi^{op} $$
Firstly, I don't see how the left or right is a functor so how can we speak of the natural map?
Secondly, how do you make the map? If we have a natural map $a in limlimits_{longleftarrow} beta$ such that $beta(f) circ a_i = a_j$ for any $f : j to i$ in $I$, I'm not seeing how this should work.
category-theory limits-colimits functors hom-functor
category-theory limits-colimits functors hom-functor
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
edited Jun 12 at 17:47
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
asked Jun 12 at 14:30
Roll up and smoke Adjoint
8,96752357
8,96752357
Surely there must be additional conditions on $phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton.
– Arnaud D.
Jun 12 at 14:36
1
Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $Xmapsto lim mathrm{hom}(X,beta)$ as $lim beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit.
– asdq
Jun 12 at 14:51
It's on page 36.
– Roll up and smoke Adjoint
Jun 12 at 14:53
On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $varprojlimbeta$ a functor of $beta$, in other words $varprojlim:operatorname{Fun}(I,C)to C$. Similarly if $C$ has all $J$ limits, you have a functor $operatorname{Fun}(I,C)to C$ such that $betamapstovarprojlimbetacircvarphi^{op}$. The claim is that this is a natural transformation.
– Roland
Jun 12 at 17:12
@Roland I see that part now, thank you. How do I show that there is a map between the two though?
– Roll up and smoke Adjoint
Jun 12 at 17:49
|
show 1 more comment
Surely there must be additional conditions on $phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton.
– Arnaud D.
Jun 12 at 14:36
1
Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $Xmapsto lim mathrm{hom}(X,beta)$ as $lim beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit.
– asdq
Jun 12 at 14:51
It's on page 36.
– Roll up and smoke Adjoint
Jun 12 at 14:53
On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $varprojlimbeta$ a functor of $beta$, in other words $varprojlim:operatorname{Fun}(I,C)to C$. Similarly if $C$ has all $J$ limits, you have a functor $operatorname{Fun}(I,C)to C$ such that $betamapstovarprojlimbetacircvarphi^{op}$. The claim is that this is a natural transformation.
– Roland
Jun 12 at 17:12
@Roland I see that part now, thank you. How do I show that there is a map between the two though?
– Roll up and smoke Adjoint
Jun 12 at 17:49
Surely there must be additional conditions on $phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton.
– Arnaud D.
Jun 12 at 14:36
Surely there must be additional conditions on $phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton.
– Arnaud D.
Jun 12 at 14:36
1
1
Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $Xmapsto lim mathrm{hom}(X,beta)$ as $lim beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit.
– asdq
Jun 12 at 14:51
Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $Xmapsto lim mathrm{hom}(X,beta)$ as $lim beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit.
– asdq
Jun 12 at 14:51
It's on page 36.
– Roll up and smoke Adjoint
Jun 12 at 14:53
It's on page 36.
– Roll up and smoke Adjoint
Jun 12 at 14:53
On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $varprojlimbeta$ a functor of $beta$, in other words $varprojlim:operatorname{Fun}(I,C)to C$. Similarly if $C$ has all $J$ limits, you have a functor $operatorname{Fun}(I,C)to C$ such that $betamapstovarprojlimbetacircvarphi^{op}$. The claim is that this is a natural transformation.
– Roland
Jun 12 at 17:12
On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $varprojlimbeta$ a functor of $beta$, in other words $varprojlim:operatorname{Fun}(I,C)to C$. Similarly if $C$ has all $J$ limits, you have a functor $operatorname{Fun}(I,C)to C$ such that $betamapstovarprojlimbetacircvarphi^{op}$. The claim is that this is a natural transformation.
– Roland
Jun 12 at 17:12
@Roland I see that part now, thank you. How do I show that there is a map between the two though?
– Roll up and smoke Adjoint
Jun 12 at 17:49
@Roland I see that part now, thank you. How do I show that there is a map between the two though?
– Roll up and smoke Adjoint
Jun 12 at 17:49
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let's dispense with all the duals and the values taken in a particular category, which are obscuring the situation a bit. If $D:Jto C$, $u:Ito J$ are functors such that $C$ admits limits of shapes $I$ and $J$, the claim is that we have a natural map $mathrm{lim} Dto mathrm{lim}(Dcirc u)$. For any $iin I$, we must give a map $mathrm{lim} Dto D(u(i))$.
Now, along with $mathrm{lim} D$ we are given a cone $lambda$ with components $lambda_j:mathrm{lim} Dto D(j)$. So the natural guess for the desired map is $lambda_{u(i)}$. To prove that these components determine a map into $mathrm{lim}(Dcirc u)$ as desired, we have only to check that $(Dcirc u)(f)circ lambda_{u(i)}=lambda_{u(i')}$ for every $f:ito i'$ in $I$. But this is simply an instantiation of the assumption that $lambda$ forms a cone.
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
add a comment |
up vote
0
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This proof uses just the definition of $limlimits_{leftarrow} beta = text{Hom}_{I^{wedge}}(text{pt}, beta)$ and the fact that $beta(f) circ theta_y = theta_x$ for all natural maps $theta$ in the limit, and $f : x to y$ in $I$.
Suppose also that since $varphi(g) in I^{op}, forall g in J^{op}$, we have $varphi(g): varphi(b) to varphi(a)$ for all $g in J^{op}$ so that by letting $f = varphi(g)$ in the above we get that $beta(varphi(g)) circ theta_{varphi(b)} = theta_{varphi(a)}$. Thus for any $theta$ in the first limit, there is $theta circ varphi equiv theta_{varphi(cdot)}$ which is a natural map in $text{Hom}_{J^{wedge}}(text{pt}, beta circ varphi)$.
answered Nov 24 at 0:29
Roll up and smoke Adjoint
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let's dispense with all the duals and the values taken in a particular category, which are obscuring the situation a bit. If $D:Jto C$, $u:Ito J$ are functors such that $C$ admits limits of shapes $I$ and $J$, the claim is that we have a natural map $mathrm{lim} Dto mathrm{lim}(Dcirc u)$. For any $iin I$, we must give a map $mathrm{lim} Dto D(u(i))$.
Now, along with $mathrm{lim} D$ we are given a cone $lambda$ with components $lambda_j:mathrm{lim} Dto D(j)$. So the natural guess for the desired map is $lambda_{u(i)}$. To prove that these components determine a map into $mathrm{lim}(Dcirc u)$ as desired, we have only to check that $(Dcirc u)(f)circ lambda_{u(i)}=lambda_{u(i')}$ for every $f:ito i'$ in $I$. But this is simply an instantiation of the assumption that $lambda$ forms a cone.
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
add a comment |
up vote
3
down vote
accepted
Let's dispense with all the duals and the values taken in a particular category, which are obscuring the situation a bit. If $D:Jto C$, $u:Ito J$ are functors such that $C$ admits limits of shapes $I$ and $J$, the claim is that we have a natural map $mathrm{lim} Dto mathrm{lim}(Dcirc u)$. For any $iin I$, we must give a map $mathrm{lim} Dto D(u(i))$.
Now, along with $mathrm{lim} D$ we are given a cone $lambda$ with components $lambda_j:mathrm{lim} Dto D(j)$. So the natural guess for the desired map is $lambda_{u(i)}$. To prove that these components determine a map into $mathrm{lim}(Dcirc u)$ as desired, we have only to check that $(Dcirc u)(f)circ lambda_{u(i)}=lambda_{u(i')}$ for every $f:ito i'$ in $I$. But this is simply an instantiation of the assumption that $lambda$ forms a cone.
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let's dispense with all the duals and the values taken in a particular category, which are obscuring the situation a bit. If $D:Jto C$, $u:Ito J$ are functors such that $C$ admits limits of shapes $I$ and $J$, the claim is that we have a natural map $mathrm{lim} Dto mathrm{lim}(Dcirc u)$. For any $iin I$, we must give a map $mathrm{lim} Dto D(u(i))$.
Now, along with $mathrm{lim} D$ we are given a cone $lambda$ with components $lambda_j:mathrm{lim} Dto D(j)$. So the natural guess for the desired map is $lambda_{u(i)}$. To prove that these components determine a map into $mathrm{lim}(Dcirc u)$ as desired, we have only to check that $(Dcirc u)(f)circ lambda_{u(i)}=lambda_{u(i')}$ for every $f:ito i'$ in $I$. But this is simply an instantiation of the assumption that $lambda$ forms a cone.
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
Let's dispense with all the duals and the values taken in a particular category, which are obscuring the situation a bit. If $D:Jto C$, $u:Ito J$ are functors such that $C$ admits limits of shapes $I$ and $J$, the claim is that we have a natural map $mathrm{lim} Dto mathrm{lim}(Dcirc u)$. For any $iin I$, we must give a map $mathrm{lim} Dto D(u(i))$.
Now, along with $mathrm{lim} D$ we are given a cone $lambda$ with components $lambda_j:mathrm{lim} Dto D(j)$. So the natural guess for the desired map is $lambda_{u(i)}$. To prove that these components determine a map into $mathrm{lim}(Dcirc u)$ as desired, we have only to check that $(Dcirc u)(f)circ lambda_{u(i)}=lambda_{u(i')}$ for every $f:ito i'$ in $I$. But this is simply an instantiation of the assumption that $lambda$ forms a cone.
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
answered Jun 12 at 19:59
Kevin Carlson
32.4k23270
32.4k23270
add a comment |
add a comment |
up vote
0
down vote
This proof uses just the definition of $limlimits_{leftarrow} beta = text{Hom}_{I^{wedge}}(text{pt}, beta)$ and the fact that $beta(f) circ theta_y = theta_x$ for all natural maps $theta$ in the limit, and $f : x to y$ in $I$.
Suppose also that since $varphi(g) in I^{op}, forall g in J^{op}$, we have $varphi(g): varphi(b) to varphi(a)$ for all $g in J^{op}$ so that by letting $f = varphi(g)$ in the above we get that $beta(varphi(g)) circ theta_{varphi(b)} = theta_{varphi(a)}$. Thus for any $theta$ in the first limit, there is $theta circ varphi equiv theta_{varphi(cdot)}$ which is a natural map in $text{Hom}_{J^{wedge}}(text{pt}, beta circ varphi)$.
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
add a comment |
up vote
0
down vote
This proof uses just the definition of $limlimits_{leftarrow} beta = text{Hom}_{I^{wedge}}(text{pt}, beta)$ and the fact that $beta(f) circ theta_y = theta_x$ for all natural maps $theta$ in the limit, and $f : x to y$ in $I$.
Suppose also that since $varphi(g) in I^{op}, forall g in J^{op}$, we have $varphi(g): varphi(b) to varphi(a)$ for all $g in J^{op}$ so that by letting $f = varphi(g)$ in the above we get that $beta(varphi(g)) circ theta_{varphi(b)} = theta_{varphi(a)}$. Thus for any $theta$ in the first limit, there is $theta circ varphi equiv theta_{varphi(cdot)}$ which is a natural map in $text{Hom}_{J^{wedge}}(text{pt}, beta circ varphi)$.
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
add a comment |
up vote
0
down vote
up vote
0
down vote
This proof uses just the definition of $limlimits_{leftarrow} beta = text{Hom}_{I^{wedge}}(text{pt}, beta)$ and the fact that $beta(f) circ theta_y = theta_x$ for all natural maps $theta$ in the limit, and $f : x to y$ in $I$.
Suppose also that since $varphi(g) in I^{op}, forall g in J^{op}$, we have $varphi(g): varphi(b) to varphi(a)$ for all $g in J^{op}$ so that by letting $f = varphi(g)$ in the above we get that $beta(varphi(g)) circ theta_{varphi(b)} = theta_{varphi(a)}$. Thus for any $theta$ in the first limit, there is $theta circ varphi equiv theta_{varphi(cdot)}$ which is a natural map in $text{Hom}_{J^{wedge}}(text{pt}, beta circ varphi)$.
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
This proof uses just the definition of $limlimits_{leftarrow} beta = text{Hom}_{I^{wedge}}(text{pt}, beta)$ and the fact that $beta(f) circ theta_y = theta_x$ for all natural maps $theta$ in the limit, and $f : x to y$ in $I$.
Suppose also that since $varphi(g) in I^{op}, forall g in J^{op}$, we have $varphi(g): varphi(b) to varphi(a)$ for all $g in J^{op}$ so that by letting $f = varphi(g)$ in the above we get that $beta(varphi(g)) circ theta_{varphi(b)} = theta_{varphi(a)}$. Thus for any $theta$ in the first limit, there is $theta circ varphi equiv theta_{varphi(cdot)}$ which is a natural map in $text{Hom}_{J^{wedge}}(text{pt}, beta circ varphi)$.
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
answered Nov 24 at 0:29
Roll up and smoke Adjoint
8,96752357
8,96752357
add a comment |
add a comment |
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Surely there must be additional conditions on $phi$? This would certainly fail if, for example, $J$ was the empty category, since it would imply that every limit is the terminal object, and thus that every set is a singleton.
– Arnaud D.
Jun 12 at 14:36
1
Which exercise is this? Regarding the issue of how both sides are functors, in the book they also refer to the functor $Xmapsto lim mathrm{hom}(X,beta)$ as $lim beta$, which in the case the limit exists in the respective category is the Yoneda embedding of the limit.
– asdq
Jun 12 at 14:51
It's on page 36.
– Roll up and smoke Adjoint
Jun 12 at 14:53
On page 36, I read a natural morphism (not isomorphism). If $C$ has all $I$-limits, you can make $varprojlimbeta$ a functor of $beta$, in other words $varprojlim:operatorname{Fun}(I,C)to C$. Similarly if $C$ has all $J$ limits, you have a functor $operatorname{Fun}(I,C)to C$ such that $betamapstovarprojlimbetacircvarphi^{op}$. The claim is that this is a natural transformation.
– Roland
Jun 12 at 17:12
@Roland I see that part now, thank you. How do I show that there is a map between the two though?
– Roll up and smoke Adjoint
Jun 12 at 17:49