Mixing
Inverse Limit as a Functor, is this correct?
$begingroup$
Suppose we have a homomorphism $psi: R/I^nto S/J^n$.
Does it follow from the fact that inverse limit is a functor that it extends to a homomorphism
$$widetilde{psi}: varprojlim R/I^nto varprojlim S/J^n$$?
Thanks a lot.
abstract-algebra category-theory
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
$endgroup$
add a comment |
$begingroup$
Suppose we have a homomorphism $psi: R/I^nto S/J^n$.
Does it follow from the fact that inverse limit is a functor that it extends to a homomorphism
$$widetilde{psi}: varprojlim R/I^nto varprojlim S/J^n$$?
Thanks a lot.
abstract-algebra category-theory
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
$endgroup$
$begingroup$
Yes: functors send morphisms to morphisms.
$endgroup$
– user3482749
Jan 2 at 15:27
$begingroup$
Thanks. According to Wikipedia (en.wikipedia.org/wiki/Inverse_limit), the inverse limit $varprojlim: C^{I^{op}}to C$ is a functor. I am concerned about the $C^{I^{op}}$, and whether my question still holds.
$endgroup$
– yoyostein
Jan 2 at 15:31
add a comment |
$begingroup$
Suppose we have a homomorphism $psi: R/I^nto S/J^n$.
Does it follow from the fact that inverse limit is a functor that it extends to a homomorphism
$$widetilde{psi}: varprojlim R/I^nto varprojlim S/J^n$$?
Thanks a lot.
abstract-algebra category-theory
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
$endgroup$
Suppose we have a homomorphism $psi: R/I^nto S/J^n$.
Does it follow from the fact that inverse limit is a functor that it extends to a homomorphism
$$widetilde{psi}: varprojlim R/I^nto varprojlim S/J^n$$?
Thanks a lot.
abstract-algebra category-theory
abstract-algebra category-theory
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
asked Jan 2 at 15:18
yoyosteinyoyostein
7,90293768
7,90293768
$begingroup$
Yes: functors send morphisms to morphisms.
$endgroup$
– user3482749
Jan 2 at 15:27
$begingroup$
Thanks. According to Wikipedia (en.wikipedia.org/wiki/Inverse_limit), the inverse limit $varprojlim: C^{I^{op}}to C$ is a functor. I am concerned about the $C^{I^{op}}$, and whether my question still holds.
$endgroup$
– yoyostein
Jan 2 at 15:31
add a comment |
$begingroup$
Yes: functors send morphisms to morphisms.
$endgroup$
– user3482749
Jan 2 at 15:27
$begingroup$
Thanks. According to Wikipedia (en.wikipedia.org/wiki/Inverse_limit), the inverse limit $varprojlim: C^{I^{op}}to C$ is a functor. I am concerned about the $C^{I^{op}}$, and whether my question still holds.
$endgroup$
– yoyostein
Jan 2 at 15:31
$begingroup$
Yes: functors send morphisms to morphisms.
$endgroup$
– user3482749
Jan 2 at 15:27
$begingroup$
Yes: functors send morphisms to morphisms.
$endgroup$
– user3482749
Jan 2 at 15:27
$begingroup$
Thanks. According to Wikipedia (en.wikipedia.org/wiki/Inverse_limit), the inverse limit $varprojlim: C^{I^{op}}to C$ is a functor. I am concerned about the $C^{I^{op}}$, and whether my question still holds.
$endgroup$
– yoyostein
Jan 2 at 15:31
$begingroup$
Thanks. According to Wikipedia (en.wikipedia.org/wiki/Inverse_limit), the inverse limit $varprojlim: C^{I^{op}}to C$ is a functor. I am concerned about the $C^{I^{op}}$, and whether my question still holds.
$endgroup$
– yoyostein
Jan 2 at 15:31
add a comment |
1 Answer
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$begingroup$
A few things:
The inverse limit is a covariant functor from [the category of directed systems in $C$] to $C$.
The category of directed systems in $C$ has a slick description in terms of the opposite category of some category describing the indexing, which is what the Wikipedia article you're linking is doing. But you should not think the inverse limit itself is a contravariant functor; it is covariant (as the page explicitly says).
Your notation is a little ambiguous. To get a morphism of directed systems, we don't just want one $psi$, but for each $n$ we want a $psi_n: R/I^n to S/J^n$. Further, we require that these commute with the natural quotient maps $R/I^n to R/I^{n-1}$ and $S/J^n to S/J^{n-1}$. It's only with this data that we get a morphism of inverse limits (EDIT: or rather, it's only with this data that we get a morphism of inverse limits by directly appealing to functoriality -- see the comments).
answered Jan 2 at 16:07
hunterhunter
14.4k22438
$endgroup$
1
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
A few things:
The inverse limit is a covariant functor from [the category of directed systems in $C$] to $C$.
The category of directed systems in $C$ has a slick description in terms of the opposite category of some category describing the indexing, which is what the Wikipedia article you're linking is doing. But you should not think the inverse limit itself is a contravariant functor; it is covariant (as the page explicitly says).
Your notation is a little ambiguous. To get a morphism of directed systems, we don't just want one $psi$, but for each $n$ we want a $psi_n: R/I^n to S/J^n$. Further, we require that these commute with the natural quotient maps $R/I^n to R/I^{n-1}$ and $S/J^n to S/J^{n-1}$. It's only with this data that we get a morphism of inverse limits (EDIT: or rather, it's only with this data that we get a morphism of inverse limits by directly appealing to functoriality -- see the comments).
answered Jan 2 at 16:07
hunterhunter
14.4k22438
$endgroup$
1
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
add a comment |
$begingroup$
A few things:
The inverse limit is a covariant functor from [the category of directed systems in $C$] to $C$.
The category of directed systems in $C$ has a slick description in terms of the opposite category of some category describing the indexing, which is what the Wikipedia article you're linking is doing. But you should not think the inverse limit itself is a contravariant functor; it is covariant (as the page explicitly says).
Your notation is a little ambiguous. To get a morphism of directed systems, we don't just want one $psi$, but for each $n$ we want a $psi_n: R/I^n to S/J^n$. Further, we require that these commute with the natural quotient maps $R/I^n to R/I^{n-1}$ and $S/J^n to S/J^{n-1}$. It's only with this data that we get a morphism of inverse limits (EDIT: or rather, it's only with this data that we get a morphism of inverse limits by directly appealing to functoriality -- see the comments).
answered Jan 2 at 16:07
hunterhunter
14.4k22438
$endgroup$
1
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
add a comment |
$begingroup$
A few things:
The inverse limit is a covariant functor from [the category of directed systems in $C$] to $C$.
The category of directed systems in $C$ has a slick description in terms of the opposite category of some category describing the indexing, which is what the Wikipedia article you're linking is doing. But you should not think the inverse limit itself is a contravariant functor; it is covariant (as the page explicitly says).
Your notation is a little ambiguous. To get a morphism of directed systems, we don't just want one $psi$, but for each $n$ we want a $psi_n: R/I^n to S/J^n$. Further, we require that these commute with the natural quotient maps $R/I^n to R/I^{n-1}$ and $S/J^n to S/J^{n-1}$. It's only with this data that we get a morphism of inverse limits (EDIT: or rather, it's only with this data that we get a morphism of inverse limits by directly appealing to functoriality -- see the comments).
answered Jan 2 at 16:07
hunterhunter
14.4k22438
$endgroup$
A few things:
The inverse limit is a covariant functor from [the category of directed systems in $C$] to $C$.
The category of directed systems in $C$ has a slick description in terms of the opposite category of some category describing the indexing, which is what the Wikipedia article you're linking is doing. But you should not think the inverse limit itself is a contravariant functor; it is covariant (as the page explicitly says).
Your notation is a little ambiguous. To get a morphism of directed systems, we don't just want one $psi$, but for each $n$ we want a $psi_n: R/I^n to S/J^n$. Further, we require that these commute with the natural quotient maps $R/I^n to R/I^{n-1}$ and $S/J^n to S/J^{n-1}$. It's only with this data that we get a morphism of inverse limits (EDIT: or rather, it's only with this data that we get a morphism of inverse limits by directly appealing to functoriality -- see the comments).
answered Jan 2 at 16:07
hunterhunter
14.4k22438
edited Jan 2 at 16:25
answered Jan 2 at 16:07
hunterhunter
14.4k22438
answered Jan 2 at 16:07
hunterhunter
14.4k22438
answered Jan 2 at 16:07
hunterhunter
14.4k22438
14.4k22438
1
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
add a comment |
1
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
1
1
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
The last sentence of this answer should probably say that it's only from these data that we get a morphism of inverse limits directly from functoriality. Other situations might get us a morphism of inverse limits via some additional work.
$endgroup$
– Andreas Blass
Jan 2 at 16:16
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
$begingroup$
@AndreasBlass good point! I'll edit it into the answer.
$endgroup$
– hunter
Jan 2 at 16:25
add a comment |
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$begingroup$
Yes: functors send morphisms to morphisms.
$endgroup$
– user3482749
Jan 2 at 15:27
$begingroup$
Thanks. According to Wikipedia (en.wikipedia.org/wiki/Inverse_limit), the inverse limit $varprojlim: C^{I^{op}}to C$ is a functor. I am concerned about the $C^{I^{op}}$, and whether my question still holds.
$endgroup$
– yoyostein
Jan 2 at 15:31