Exact functors and derived functorsDerived FunctorsExact contravariant functors and splittingWhy are functors...
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Exact functors and derived functors
Derived FunctorsExact contravariant functors and splittingWhy are functors exact if they preserve all short exact sequences?Derived functors - homotopical vs homological approachDerived functors of abelianizationDerived Functors of a Half-Exact FunctorA sequence of functor comes from derived functorDerived functor of exact functorDerived functors and induced functorsSnake lemma for derived functors
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$begingroup$
Given an exact (additive) functor $F$, i.e. an additive functor preserving exact sequences, it is not hard to show that all derived functors of $F$ vanish.
At the same time given a right exact functor (a similar argument holds for the left exact case) one can show that for every short exact sequence $$0longrightarrow Alongrightarrow Blongrightarrow Clongrightarrow 0$$ there exists an induced long exact sequence of the form $$cdots longrightarrow L_1F(B)longrightarrow L_1F(C)longrightarrow F(A) longrightarrow F(B)longrightarrow F(C) longrightarrow 0$$ and hence if the first derived functor $L_1F$ vanishes, the functor is exact.
This seems to imply that the vanishing of the first derived functor is a sufficient condition for the vanishing of all higher derived functors. Is this true? This feels like a very strong result/constraint, so I get the feeling that I am missing something.
category-theory homological-algebra exact-sequence derived-functors functors
asked 10 hours ago
NDewolfNDewolf
612311
$endgroup$
add a comment |
$begingroup$
Given an exact (additive) functor $F$, i.e. an additive functor preserving exact sequences, it is not hard to show that all derived functors of $F$ vanish.
At the same time given a right exact functor (a similar argument holds for the left exact case) one can show that for every short exact sequence $$0longrightarrow Alongrightarrow Blongrightarrow Clongrightarrow 0$$ there exists an induced long exact sequence of the form $$cdots longrightarrow L_1F(B)longrightarrow L_1F(C)longrightarrow F(A) longrightarrow F(B)longrightarrow F(C) longrightarrow 0$$ and hence if the first derived functor $L_1F$ vanishes, the functor is exact.
This seems to imply that the vanishing of the first derived functor is a sufficient condition for the vanishing of all higher derived functors. Is this true? This feels like a very strong result/constraint, so I get the feeling that I am missing something.
category-theory homological-algebra exact-sequence derived-functors functors
asked 10 hours ago
NDewolfNDewolf
612311
$endgroup$
add a comment |
$begingroup$
Given an exact (additive) functor $F$, i.e. an additive functor preserving exact sequences, it is not hard to show that all derived functors of $F$ vanish.
At the same time given a right exact functor (a similar argument holds for the left exact case) one can show that for every short exact sequence $$0longrightarrow Alongrightarrow Blongrightarrow Clongrightarrow 0$$ there exists an induced long exact sequence of the form $$cdots longrightarrow L_1F(B)longrightarrow L_1F(C)longrightarrow F(A) longrightarrow F(B)longrightarrow F(C) longrightarrow 0$$ and hence if the first derived functor $L_1F$ vanishes, the functor is exact.
This seems to imply that the vanishing of the first derived functor is a sufficient condition for the vanishing of all higher derived functors. Is this true? This feels like a very strong result/constraint, so I get the feeling that I am missing something.
category-theory homological-algebra exact-sequence derived-functors functors
asked 10 hours ago
NDewolfNDewolf
612311
$endgroup$
Given an exact (additive) functor $F$, i.e. an additive functor preserving exact sequences, it is not hard to show that all derived functors of $F$ vanish.
At the same time given a right exact functor (a similar argument holds for the left exact case) one can show that for every short exact sequence $$0longrightarrow Alongrightarrow Blongrightarrow Clongrightarrow 0$$ there exists an induced long exact sequence of the form $$cdots longrightarrow L_1F(B)longrightarrow L_1F(C)longrightarrow F(A) longrightarrow F(B)longrightarrow F(C) longrightarrow 0$$ and hence if the first derived functor $L_1F$ vanishes, the functor is exact.
This seems to imply that the vanishing of the first derived functor is a sufficient condition for the vanishing of all higher derived functors. Is this true? This feels like a very strong result/constraint, so I get the feeling that I am missing something.
category-theory homological-algebra exact-sequence derived-functors functors
category-theory homological-algebra exact-sequence derived-functors functors
asked 10 hours ago
NDewolfNDewolf
612311
asked 10 hours ago
NDewolfNDewolf
612311
asked 10 hours ago
NDewolfNDewolf
612311
asked 10 hours ago
NDewolfNDewolf
612311
asked 10 hours ago
NDewolfNDewolf
612311
612311
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects.
The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0to Bto P to Ato 0$$ where $P$ is projective. There is then an induced long exact sequence $$dotsto L_{n+1}F(P)to L_{n+1}F(A)to L_nF(B)to L_nF(P)tocdots$$
When $ngeq 1$, $L_nF(P)$ and $L_{n+1}F(P)$ are trivial since $P$ is projective, and so the map $L_{n+1}F(A)to L_nF(B)$ is an isomorphism. So, for instance, the vanishing of $L_1F(B)$ is equivalent to the vanishing of $L_2F(A)$. Iterating this construction, we can similarly find an object $C$ such that the vanishing of $L_1F(C)$ is equivalent to the vanishing of $L_3F(A)$, and so on. So if we know $L_1F$ vanishes on all objects, that actually tells us $L_nF$ vanishes on $A$ for all $ngeq 1$.
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
$endgroup$
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
add a comment |
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$begingroup$
Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects.
The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0to Bto P to Ato 0$$ where $P$ is projective. There is then an induced long exact sequence $$dotsto L_{n+1}F(P)to L_{n+1}F(A)to L_nF(B)to L_nF(P)tocdots$$
When $ngeq 1$, $L_nF(P)$ and $L_{n+1}F(P)$ are trivial since $P$ is projective, and so the map $L_{n+1}F(A)to L_nF(B)$ is an isomorphism. So, for instance, the vanishing of $L_1F(B)$ is equivalent to the vanishing of $L_2F(A)$. Iterating this construction, we can similarly find an object $C$ such that the vanishing of $L_1F(C)$ is equivalent to the vanishing of $L_3F(A)$, and so on. So if we know $L_1F$ vanishes on all objects, that actually tells us $L_nF$ vanishes on $A$ for all $ngeq 1$.
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
$endgroup$
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
add a comment |
$begingroup$
Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects.
The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0to Bto P to Ato 0$$ where $P$ is projective. There is then an induced long exact sequence $$dotsto L_{n+1}F(P)to L_{n+1}F(A)to L_nF(B)to L_nF(P)tocdots$$
When $ngeq 1$, $L_nF(P)$ and $L_{n+1}F(P)$ are trivial since $P$ is projective, and so the map $L_{n+1}F(A)to L_nF(B)$ is an isomorphism. So, for instance, the vanishing of $L_1F(B)$ is equivalent to the vanishing of $L_2F(A)$. Iterating this construction, we can similarly find an object $C$ such that the vanishing of $L_1F(C)$ is equivalent to the vanishing of $L_3F(A)$, and so on. So if we know $L_1F$ vanishes on all objects, that actually tells us $L_nF$ vanishes on $A$ for all $ngeq 1$.
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
$endgroup$
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
add a comment |
$begingroup$
Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects.
The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0to Bto P to Ato 0$$ where $P$ is projective. There is then an induced long exact sequence $$dotsto L_{n+1}F(P)to L_{n+1}F(A)to L_nF(B)to L_nF(P)tocdots$$
When $ngeq 1$, $L_nF(P)$ and $L_{n+1}F(P)$ are trivial since $P$ is projective, and so the map $L_{n+1}F(A)to L_nF(B)$ is an isomorphism. So, for instance, the vanishing of $L_1F(B)$ is equivalent to the vanishing of $L_2F(A)$. Iterating this construction, we can similarly find an object $C$ such that the vanishing of $L_1F(C)$ is equivalent to the vanishing of $L_3F(A)$, and so on. So if we know $L_1F$ vanishes on all objects, that actually tells us $L_nF$ vanishes on $A$ for all $ngeq 1$.
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
$endgroup$
Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects.
The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0to Bto P to Ato 0$$ where $P$ is projective. There is then an induced long exact sequence $$dotsto L_{n+1}F(P)to L_{n+1}F(A)to L_nF(B)to L_nF(P)tocdots$$
When $ngeq 1$, $L_nF(P)$ and $L_{n+1}F(P)$ are trivial since $P$ is projective, and so the map $L_{n+1}F(A)to L_nF(B)$ is an isomorphism. So, for instance, the vanishing of $L_1F(B)$ is equivalent to the vanishing of $L_2F(A)$. Iterating this construction, we can similarly find an object $C$ such that the vanishing of $L_1F(C)$ is equivalent to the vanishing of $L_3F(A)$, and so on. So if we know $L_1F$ vanishes on all objects, that actually tells us $L_nF$ vanishes on $A$ for all $ngeq 1$.
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
edited 10 hours ago
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
answered 10 hours ago
Eric WofseyEric Wofsey
200k14231362
200k14231362
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
add a comment |
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
Thanks for the quick and very informative answer. This construction indeed gives a lot of insight.
$endgroup$
– NDewolf
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
$begingroup$
This trick is known as "dimension-shifting".
$endgroup$
– Lord Shark the Unknown
10 hours ago
add a comment |
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