Mixing
About category theory and direct sum
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What is the explicit meaning of the statement that * is compatible with direct sums ; what are these maps ; I guess they are projection and injection map but could somebody explain how duality works in this kind of maps intutively? Also how Hom(M,N) is isomorphic to Hom(M**,N**)?
abstract-algebra ring-theory modules
asked Feb 1 at 18:37
maths studentmaths student
6321521
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add a comment |
$begingroup$
What is the explicit meaning of the statement that * is compatible with direct sums ; what are these maps ; I guess they are projection and injection map but could somebody explain how duality works in this kind of maps intutively? Also how Hom(M,N) is isomorphic to Hom(M**,N**)?
abstract-algebra ring-theory modules
asked Feb 1 at 18:37
maths studentmaths student
6321521
$endgroup$
add a comment |
$begingroup$
What is the explicit meaning of the statement that * is compatible with direct sums ; what are these maps ; I guess they are projection and injection map but could somebody explain how duality works in this kind of maps intutively? Also how Hom(M,N) is isomorphic to Hom(M**,N**)?
abstract-algebra ring-theory modules
asked Feb 1 at 18:37
maths studentmaths student
6321521
$endgroup$
What is the explicit meaning of the statement that * is compatible with direct sums ; what are these maps ; I guess they are projection and injection map but could somebody explain how duality works in this kind of maps intutively? Also how Hom(M,N) is isomorphic to Hom(M**,N**)?
abstract-algebra ring-theory modules
abstract-algebra ring-theory modules
asked Feb 1 at 18:37
maths studentmaths student
6321521
asked Feb 1 at 18:37
maths studentmaths student
6321521
edited Feb 6 at 6:58
maths student
asked Feb 1 at 18:37
maths studentmaths student
6321521
asked Feb 1 at 18:37
maths studentmaths student
6321521
asked Feb 1 at 18:37
maths studentmaths student
6321521
6321521
add a comment |
add a comment |
2 Answers
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oldest
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Well as is said $*$ is a functor, let's say $Mod_R to Mod_R$ for some commutative ring $R$. The compatibility with direct sums can be rephrased categorically as does this duality functor preserve finite coproducts (which are the same as finite products in an additive category). The projection and injection map are the maps defined for the product/coproduct respectively. For your final question, if we know that $**$ is naturally isomorphic to the identity functor, then in particular it is full and faithful, which means that we get an isomorphism of Hom sets.
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
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add a comment |
$begingroup$
$*$ is a contravariant functor $mathscr{Mto M}$.
So if you have a canonical projection map $pi_i : M_1oplus ...oplus M_n to M_i$, you get $pi_i^* : M_i^*to (M_1oplus ...oplus M_n)^*$. The requirement is that this is the "injection" of $M_i^*$ into a direct sum of the $M_i^*$'s.
Dually, if you have a canonical injection $rho_i : M_i to M_1oplus ...oplus M_n$, the requirement is that $rho_i^*$ be the canonical projection map (note that the author says "direct sum" whereas after they only mention finite direct sums; this is something that should be clarified)
For your last question, there's a requirement that $(-)^{**}$ be naturally isomorphic to $id_mathscr{M}$, but classically if $Fsimeq G$, then $hom (F(x),F(y))cong hom (G(x), G(y))$. It's just the same argument as saying that if $Asimeq B, Csimeq D$, then $hom (A,C) simeq hom (B, D)$, only that since the initial isomorphism are natural, so are the induced ones.
Note that this doesn't follow from the other assumptions, it is another assumption
answered Feb 1 at 20:24
MaxMax
16.1k11144
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2 Answers
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active
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2 Answers
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$begingroup$
Well as is said $*$ is a functor, let's say $Mod_R to Mod_R$ for some commutative ring $R$. The compatibility with direct sums can be rephrased categorically as does this duality functor preserve finite coproducts (which are the same as finite products in an additive category). The projection and injection map are the maps defined for the product/coproduct respectively. For your final question, if we know that $**$ is naturally isomorphic to the identity functor, then in particular it is full and faithful, which means that we get an isomorphism of Hom sets.
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
$endgroup$
add a comment |
$begingroup$
Well as is said $*$ is a functor, let's say $Mod_R to Mod_R$ for some commutative ring $R$. The compatibility with direct sums can be rephrased categorically as does this duality functor preserve finite coproducts (which are the same as finite products in an additive category). The projection and injection map are the maps defined for the product/coproduct respectively. For your final question, if we know that $**$ is naturally isomorphic to the identity functor, then in particular it is full and faithful, which means that we get an isomorphism of Hom sets.
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
$endgroup$
add a comment |
$begingroup$
Well as is said $*$ is a functor, let's say $Mod_R to Mod_R$ for some commutative ring $R$. The compatibility with direct sums can be rephrased categorically as does this duality functor preserve finite coproducts (which are the same as finite products in an additive category). The projection and injection map are the maps defined for the product/coproduct respectively. For your final question, if we know that $**$ is naturally isomorphic to the identity functor, then in particular it is full and faithful, which means that we get an isomorphism of Hom sets.
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
$endgroup$
Well as is said $*$ is a functor, let's say $Mod_R to Mod_R$ for some commutative ring $R$. The compatibility with direct sums can be rephrased categorically as does this duality functor preserve finite coproducts (which are the same as finite products in an additive category). The projection and injection map are the maps defined for the product/coproduct respectively. For your final question, if we know that $**$ is naturally isomorphic to the identity functor, then in particular it is full and faithful, which means that we get an isomorphism of Hom sets.
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
answered Feb 1 at 18:55
Sheel StueberSheel Stueber
691210
691210
add a comment |
add a comment |
$begingroup$
$*$ is a contravariant functor $mathscr{Mto M}$.
So if you have a canonical projection map $pi_i : M_1oplus ...oplus M_n to M_i$, you get $pi_i^* : M_i^*to (M_1oplus ...oplus M_n)^*$. The requirement is that this is the "injection" of $M_i^*$ into a direct sum of the $M_i^*$'s.
Dually, if you have a canonical injection $rho_i : M_i to M_1oplus ...oplus M_n$, the requirement is that $rho_i^*$ be the canonical projection map (note that the author says "direct sum" whereas after they only mention finite direct sums; this is something that should be clarified)
For your last question, there's a requirement that $(-)^{**}$ be naturally isomorphic to $id_mathscr{M}$, but classically if $Fsimeq G$, then $hom (F(x),F(y))cong hom (G(x), G(y))$. It's just the same argument as saying that if $Asimeq B, Csimeq D$, then $hom (A,C) simeq hom (B, D)$, only that since the initial isomorphism are natural, so are the induced ones.
Note that this doesn't follow from the other assumptions, it is another assumption
answered Feb 1 at 20:24
MaxMax
16.1k11144
$endgroup$
add a comment |
$begingroup$
$*$ is a contravariant functor $mathscr{Mto M}$.
So if you have a canonical projection map $pi_i : M_1oplus ...oplus M_n to M_i$, you get $pi_i^* : M_i^*to (M_1oplus ...oplus M_n)^*$. The requirement is that this is the "injection" of $M_i^*$ into a direct sum of the $M_i^*$'s.
Dually, if you have a canonical injection $rho_i : M_i to M_1oplus ...oplus M_n$, the requirement is that $rho_i^*$ be the canonical projection map (note that the author says "direct sum" whereas after they only mention finite direct sums; this is something that should be clarified)
For your last question, there's a requirement that $(-)^{**}$ be naturally isomorphic to $id_mathscr{M}$, but classically if $Fsimeq G$, then $hom (F(x),F(y))cong hom (G(x), G(y))$. It's just the same argument as saying that if $Asimeq B, Csimeq D$, then $hom (A,C) simeq hom (B, D)$, only that since the initial isomorphism are natural, so are the induced ones.
Note that this doesn't follow from the other assumptions, it is another assumption
answered Feb 1 at 20:24
MaxMax
16.1k11144
$endgroup$
add a comment |
$begingroup$
$*$ is a contravariant functor $mathscr{Mto M}$.
So if you have a canonical projection map $pi_i : M_1oplus ...oplus M_n to M_i$, you get $pi_i^* : M_i^*to (M_1oplus ...oplus M_n)^*$. The requirement is that this is the "injection" of $M_i^*$ into a direct sum of the $M_i^*$'s.
Dually, if you have a canonical injection $rho_i : M_i to M_1oplus ...oplus M_n$, the requirement is that $rho_i^*$ be the canonical projection map (note that the author says "direct sum" whereas after they only mention finite direct sums; this is something that should be clarified)
For your last question, there's a requirement that $(-)^{**}$ be naturally isomorphic to $id_mathscr{M}$, but classically if $Fsimeq G$, then $hom (F(x),F(y))cong hom (G(x), G(y))$. It's just the same argument as saying that if $Asimeq B, Csimeq D$, then $hom (A,C) simeq hom (B, D)$, only that since the initial isomorphism are natural, so are the induced ones.
Note that this doesn't follow from the other assumptions, it is another assumption
answered Feb 1 at 20:24
MaxMax
16.1k11144
$endgroup$
$*$ is a contravariant functor $mathscr{Mto M}$.
So if you have a canonical projection map $pi_i : M_1oplus ...oplus M_n to M_i$, you get $pi_i^* : M_i^*to (M_1oplus ...oplus M_n)^*$. The requirement is that this is the "injection" of $M_i^*$ into a direct sum of the $M_i^*$'s.
Dually, if you have a canonical injection $rho_i : M_i to M_1oplus ...oplus M_n$, the requirement is that $rho_i^*$ be the canonical projection map (note that the author says "direct sum" whereas after they only mention finite direct sums; this is something that should be clarified)
For your last question, there's a requirement that $(-)^{**}$ be naturally isomorphic to $id_mathscr{M}$, but classically if $Fsimeq G$, then $hom (F(x),F(y))cong hom (G(x), G(y))$. It's just the same argument as saying that if $Asimeq B, Csimeq D$, then $hom (A,C) simeq hom (B, D)$, only that since the initial isomorphism are natural, so are the induced ones.
Note that this doesn't follow from the other assumptions, it is another assumption
answered Feb 1 at 20:24
MaxMax
16.1k11144
answered Feb 1 at 20:24
MaxMax
16.1k11144
answered Feb 1 at 20:24
MaxMax
16.1k11144
answered Feb 1 at 20:24
MaxMax
16.1k11144
16.1k11144
add a comment |
add a comment |
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