$Hom_{Ab}(A,-)$ is not strongly additive for some $Ain Obj(Ab)$
$begingroup$
This is an exercise question in Dold Algebraic Topology, Chpt VI, 2.12. The point is to demonstrate that additive functors does not preserve infinite direct sums.
$textbf{Q:}$ What is the example showing that $Hom_{Ab}(A,-)$ does not preserving infinite direct sums? I tried $A=Z_4, Z_2$ with $-=oplus Z_4, oplus Z_2$. However, it seems I cannot get something uncountable out by taking a countable finite direct sums. Any hint will be helpful.
abstract-algebra algebraic-topology
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
$endgroup$
add a comment |
$begingroup$
This is an exercise question in Dold Algebraic Topology, Chpt VI, 2.12. The point is to demonstrate that additive functors does not preserve infinite direct sums.
$textbf{Q:}$ What is the example showing that $Hom_{Ab}(A,-)$ does not preserving infinite direct sums? I tried $A=Z_4, Z_2$ with $-=oplus Z_4, oplus Z_2$. However, it seems I cannot get something uncountable out by taking a countable finite direct sums. Any hint will be helpful.
abstract-algebra algebraic-topology
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
$endgroup$
2
$begingroup$
You need to take $A$ to be infinitely generated.
$endgroup$
– Qiaochu Yuan
Dec 24 '18 at 4:36
$begingroup$
@QiaochuYuan Maybe this is a dumb question. I followed your hint to consider $oplus Hom(oplus Z,Z)congoplusprod Hom(Z,Z)to Hom(oplus Z,oplus Z)congprod Hom(Z,oplus Z)$. So this boils down to $oplusprod Zto prodoplus Z$ is not surjection. I think the essential point is to see $oplus Hom(oplus Z, Z)to Hom(oplus Z,oplus Z)$ is not surjection by following. There is more elements in $Hom(oplus Z,oplus Z)$.(Every element of $oplus Z$ is finite combintation but each $f$'s image being finite sum does not imply $f$ is finite sum of functions of $Hom(oplus Z,Z)$.)
$endgroup$
– user45765
Dec 24 '18 at 16:56
$begingroup$
@QiaochuYuan Is above idea correct? Thank.
$endgroup$
– user45765
Dec 24 '18 at 16:57
add a comment |
$begingroup$
This is an exercise question in Dold Algebraic Topology, Chpt VI, 2.12. The point is to demonstrate that additive functors does not preserve infinite direct sums.
$textbf{Q:}$ What is the example showing that $Hom_{Ab}(A,-)$ does not preserving infinite direct sums? I tried $A=Z_4, Z_2$ with $-=oplus Z_4, oplus Z_2$. However, it seems I cannot get something uncountable out by taking a countable finite direct sums. Any hint will be helpful.
abstract-algebra algebraic-topology
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
$endgroup$
This is an exercise question in Dold Algebraic Topology, Chpt VI, 2.12. The point is to demonstrate that additive functors does not preserve infinite direct sums.
$textbf{Q:}$ What is the example showing that $Hom_{Ab}(A,-)$ does not preserving infinite direct sums? I tried $A=Z_4, Z_2$ with $-=oplus Z_4, oplus Z_2$. However, it seems I cannot get something uncountable out by taking a countable finite direct sums. Any hint will be helpful.
abstract-algebra algebraic-topology
abstract-algebra algebraic-topology
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
edited Dec 24 '18 at 6:53
Henno Brandsma
110k347117
110k347117
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
asked Dec 24 '18 at 1:33
user45765user45765
2,6152724
2,6152724
2
$begingroup$
You need to take $A$ to be infinitely generated.
$endgroup$
– Qiaochu Yuan
Dec 24 '18 at 4:36
$begingroup$
@QiaochuYuan Maybe this is a dumb question. I followed your hint to consider $oplus Hom(oplus Z,Z)congoplusprod Hom(Z,Z)to Hom(oplus Z,oplus Z)congprod Hom(Z,oplus Z)$. So this boils down to $oplusprod Zto prodoplus Z$ is not surjection. I think the essential point is to see $oplus Hom(oplus Z, Z)to Hom(oplus Z,oplus Z)$ is not surjection by following. There is more elements in $Hom(oplus Z,oplus Z)$.(Every element of $oplus Z$ is finite combintation but each $f$'s image being finite sum does not imply $f$ is finite sum of functions of $Hom(oplus Z,Z)$.)
$endgroup$
– user45765
Dec 24 '18 at 16:56
$begingroup$
@QiaochuYuan Is above idea correct? Thank.
$endgroup$
– user45765
Dec 24 '18 at 16:57
add a comment |
2
$begingroup$
You need to take $A$ to be infinitely generated.
$endgroup$
– Qiaochu Yuan
Dec 24 '18 at 4:36
$begingroup$
@QiaochuYuan Maybe this is a dumb question. I followed your hint to consider $oplus Hom(oplus Z,Z)congoplusprod Hom(Z,Z)to Hom(oplus Z,oplus Z)congprod Hom(Z,oplus Z)$. So this boils down to $oplusprod Zto prodoplus Z$ is not surjection. I think the essential point is to see $oplus Hom(oplus Z, Z)to Hom(oplus Z,oplus Z)$ is not surjection by following. There is more elements in $Hom(oplus Z,oplus Z)$.(Every element of $oplus Z$ is finite combintation but each $f$'s image being finite sum does not imply $f$ is finite sum of functions of $Hom(oplus Z,Z)$.)
$endgroup$
– user45765
Dec 24 '18 at 16:56
$begingroup$
@QiaochuYuan Is above idea correct? Thank.
$endgroup$
– user45765
Dec 24 '18 at 16:57
2
2
$begingroup$
You need to take $A$ to be infinitely generated.
$endgroup$
– Qiaochu Yuan
Dec 24 '18 at 4:36
$begingroup$
You need to take $A$ to be infinitely generated.
$endgroup$
– Qiaochu Yuan
Dec 24 '18 at 4:36
$begingroup$
@QiaochuYuan Maybe this is a dumb question. I followed your hint to consider $oplus Hom(oplus Z,Z)congoplusprod Hom(Z,Z)to Hom(oplus Z,oplus Z)congprod Hom(Z,oplus Z)$. So this boils down to $oplusprod Zto prodoplus Z$ is not surjection. I think the essential point is to see $oplus Hom(oplus Z, Z)to Hom(oplus Z,oplus Z)$ is not surjection by following. There is more elements in $Hom(oplus Z,oplus Z)$.(Every element of $oplus Z$ is finite combintation but each $f$'s image being finite sum does not imply $f$ is finite sum of functions of $Hom(oplus Z,Z)$.)
$endgroup$
– user45765
Dec 24 '18 at 16:56
$begingroup$
@QiaochuYuan Maybe this is a dumb question. I followed your hint to consider $oplus Hom(oplus Z,Z)congoplusprod Hom(Z,Z)to Hom(oplus Z,oplus Z)congprod Hom(Z,oplus Z)$. So this boils down to $oplusprod Zto prodoplus Z$ is not surjection. I think the essential point is to see $oplus Hom(oplus Z, Z)to Hom(oplus Z,oplus Z)$ is not surjection by following. There is more elements in $Hom(oplus Z,oplus Z)$.(Every element of $oplus Z$ is finite combintation but each $f$'s image being finite sum does not imply $f$ is finite sum of functions of $Hom(oplus Z,Z)$.)
$endgroup$
– user45765
Dec 24 '18 at 16:56
$begingroup$
@QiaochuYuan Is above idea correct? Thank.
$endgroup$
– user45765
Dec 24 '18 at 16:57
$begingroup$
@QiaochuYuan Is above idea correct? Thank.
$endgroup$
– user45765
Dec 24 '18 at 16:57
add a comment |
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$begingroup$
You need to take $A$ to be infinitely generated.
$endgroup$
– Qiaochu Yuan
Dec 24 '18 at 4:36
$begingroup$
@QiaochuYuan Maybe this is a dumb question. I followed your hint to consider $oplus Hom(oplus Z,Z)congoplusprod Hom(Z,Z)to Hom(oplus Z,oplus Z)congprod Hom(Z,oplus Z)$. So this boils down to $oplusprod Zto prodoplus Z$ is not surjection. I think the essential point is to see $oplus Hom(oplus Z, Z)to Hom(oplus Z,oplus Z)$ is not surjection by following. There is more elements in $Hom(oplus Z,oplus Z)$.(Every element of $oplus Z$ is finite combintation but each $f$'s image being finite sum does not imply $f$ is finite sum of functions of $Hom(oplus Z,Z)$.)
$endgroup$
– user45765
Dec 24 '18 at 16:56
$begingroup$
@QiaochuYuan Is above idea correct? Thank.
$endgroup$
– user45765
Dec 24 '18 at 16:57