Explaination of a justification: additive functors preserve limits Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Additive functors preserve split exact sequencesshowing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)Morita contexts without tearsQuick Question on a Proof of Artin-Wedderburn TheoremAdditive exact functors preserve homology of modulesThe direct limit of morphisms and the direct limit of tensor product functors$Pcong P^ast$ iff $P$ is a f.g projective module?Modules, Direct Sums, ConfusionAdditive functors preserve direct summandsAdditive Functors preserve Null Sequences
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Explaination of a justification: additive functors preserve limits
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Additive functors preserve split exact sequencesshowing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)Morita contexts without tearsQuick Question on a Proof of Artin-Wedderburn TheoremAdditive exact functors preserve homology of modulesThe direct limit of morphisms and the direct limit of tensor product functors$Pcong P^ast$ iff $P$ is a f.g projective module?Modules, Direct Sums, ConfusionAdditive functors preserve direct summandsAdditive Functors preserve Null Sequences
$begingroup$
Lemma: For any collection $ M_i_iin I$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$$rm Hom_R(oplus_i M_i, N)cong prod_i rm Hom_R(M_i,N).$$
Proof: Additive functors preserve limits.
[Ref: this link, page 2.]
Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?
I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.
abstract-algebra modules homological-algebra
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
$endgroup$
add a comment |
$begingroup$
Lemma: For any collection $ M_i_iin I$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$$rm Hom_R(oplus_i M_i, N)cong prod_i rm Hom_R(M_i,N).$$
Proof: Additive functors preserve limits.
[Ref: this link, page 2.]
Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?
I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.
abstract-algebra modules homological-algebra
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
$endgroup$
$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21
$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32
1
$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41
$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43
add a comment |
$begingroup$
Lemma: For any collection $ M_i_iin I$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$$rm Hom_R(oplus_i M_i, N)cong prod_i rm Hom_R(M_i,N).$$
Proof: Additive functors preserve limits.
[Ref: this link, page 2.]
Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?
I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.
abstract-algebra modules homological-algebra
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
$endgroup$
Lemma: For any collection $ M_i_iin I$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$$rm Hom_R(oplus_i M_i, N)cong prod_i rm Hom_R(M_i,N).$$
Proof: Additive functors preserve limits.
[Ref: this link, page 2.]
Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?
I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.
abstract-algebra modules homological-algebra
abstract-algebra modules homological-algebra
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
edited Apr 3 at 5:30
Beginner
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
asked Apr 3 at 4:59
BeginnerBeginner
4,11311226
4,11311226
$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21
$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32
1
$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41
$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43
add a comment |
$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21
$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32
1
$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41
$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43
$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21
$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21
$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32
$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32
1
1
$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41
$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41
$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43
$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.
However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ is in the commutative $R$ case.)
The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
$endgroup$
2
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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oldest
votes
$begingroup$
If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.
However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.
However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.
However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
$endgroup$
If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.
However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
answered Apr 3 at 6:40
Eric WofseyEric Wofsey
193k14221352
193k14221352
add a comment |
add a comment |
$begingroup$
Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ is in the commutative $R$ case.)
The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
$endgroup$
2
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
add a comment |
$begingroup$
Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ is in the commutative $R$ case.)
The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
$endgroup$
2
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
add a comment |
$begingroup$
Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ is in the commutative $R$ case.)
The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
$endgroup$
Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ is in the commutative $R$ case.)
The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
edited Apr 3 at 6:53
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
answered Apr 3 at 6:29
Derek ElkinsDerek Elkins
17.7k11437
17.7k11437
2
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
add a comment |
2
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
2
2
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
$R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
$endgroup$
– Eric Wofsey
Apr 3 at 6:42
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
$begingroup$
@EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
$endgroup$
– Derek Elkins
Apr 3 at 6:55
add a comment |
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$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21
$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32
1
$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41
$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43