Showing that the direct product does not satisfy the universal property of the direct sum
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I feel intuitively that for $prod_{iin mathbb N}mathbb{Z}$, as a $mathbb{Z}$−module, and $phi_i:mathbb{Z}tomathbb{Z}$ the identity map, more than one homomorphism $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$ satisfies the condition "composition with the canonical embedding yields the identity map $phi_i:mathbb{Z}tomathbb{Z}$". But I can't myself define such homomorphisms $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$.
Could anyone suggest some appropriate functions?
abstract-algebra category-theory modules
edited Dec 17 '18 at 7:20
Andrews
1,2691421
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
$endgroup$
add a comment |
$begingroup$
I feel intuitively that for $prod_{iin mathbb N}mathbb{Z}$, as a $mathbb{Z}$−module, and $phi_i:mathbb{Z}tomathbb{Z}$ the identity map, more than one homomorphism $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$ satisfies the condition "composition with the canonical embedding yields the identity map $phi_i:mathbb{Z}tomathbb{Z}$". But I can't myself define such homomorphisms $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$.
Could anyone suggest some appropriate functions?
abstract-algebra category-theory modules
edited Dec 17 '18 at 7:20
Andrews
1,2691421
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
$endgroup$
$begingroup$
By $N$ you mean the set of natural numbers?
$endgroup$
– Daniel Fischer♦
Sep 26 '14 at 15:13
add a comment |
$begingroup$
I feel intuitively that for $prod_{iin mathbb N}mathbb{Z}$, as a $mathbb{Z}$−module, and $phi_i:mathbb{Z}tomathbb{Z}$ the identity map, more than one homomorphism $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$ satisfies the condition "composition with the canonical embedding yields the identity map $phi_i:mathbb{Z}tomathbb{Z}$". But I can't myself define such homomorphisms $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$.
Could anyone suggest some appropriate functions?
abstract-algebra category-theory modules
edited Dec 17 '18 at 7:20
Andrews
1,2691421
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
$endgroup$
I feel intuitively that for $prod_{iin mathbb N}mathbb{Z}$, as a $mathbb{Z}$−module, and $phi_i:mathbb{Z}tomathbb{Z}$ the identity map, more than one homomorphism $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$ satisfies the condition "composition with the canonical embedding yields the identity map $phi_i:mathbb{Z}tomathbb{Z}$". But I can't myself define such homomorphisms $phi:prod_{iin mathbb N}mathbb{Z}tomathbb{Z}$.
Could anyone suggest some appropriate functions?
abstract-algebra category-theory modules
abstract-algebra category-theory modules
edited Dec 17 '18 at 7:20
Andrews
1,2691421
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
edited Dec 17 '18 at 7:20
Andrews
1,2691421
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
edited Dec 17 '18 at 7:20
Andrews
1,2691421
edited Dec 17 '18 at 7:20
Andrews
1,2691421
edited Dec 17 '18 at 7:20
Andrews
1,2691421
1,2691421
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
asked Sep 26 '14 at 14:44
KeithKeith
1,4251122
1,4251122
$begingroup$
By $N$ you mean the set of natural numbers?
$endgroup$
– Daniel Fischer♦
Sep 26 '14 at 15:13
add a comment |
$begingroup$
By $N$ you mean the set of natural numbers?
$endgroup$
– Daniel Fischer♦
Sep 26 '14 at 15:13
$begingroup$
By $N$ you mean the set of natural numbers?
$endgroup$
– Daniel Fischer♦
Sep 26 '14 at 15:13
$begingroup$
By $N$ you mean the set of natural numbers?
$endgroup$
– Daniel Fischer♦
Sep 26 '14 at 15:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is a well-known result by Specker (original reference, MO/10239) that homomorphisms $mathbb{Z}^mathbb{N} to mathbb{Z}$ are determined by what they do on the $e_i=(delta_{ij})_j$ and that almost all $e_i$ are mapped to zero. In particular, there is no homomorphism mapping all $e_i mapsto 1$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
$endgroup$
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is a well-known result by Specker (original reference, MO/10239) that homomorphisms $mathbb{Z}^mathbb{N} to mathbb{Z}$ are determined by what they do on the $e_i=(delta_{ij})_j$ and that almost all $e_i$ are mapped to zero. In particular, there is no homomorphism mapping all $e_i mapsto 1$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
$endgroup$
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
add a comment |
$begingroup$
It is a well-known result by Specker (original reference, MO/10239) that homomorphisms $mathbb{Z}^mathbb{N} to mathbb{Z}$ are determined by what they do on the $e_i=(delta_{ij})_j$ and that almost all $e_i$ are mapped to zero. In particular, there is no homomorphism mapping all $e_i mapsto 1$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
$endgroup$
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
add a comment |
$begingroup$
It is a well-known result by Specker (original reference, MO/10239) that homomorphisms $mathbb{Z}^mathbb{N} to mathbb{Z}$ are determined by what they do on the $e_i=(delta_{ij})_j$ and that almost all $e_i$ are mapped to zero. In particular, there is no homomorphism mapping all $e_i mapsto 1$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
$endgroup$
It is a well-known result by Specker (original reference, MO/10239) that homomorphisms $mathbb{Z}^mathbb{N} to mathbb{Z}$ are determined by what they do on the $e_i=(delta_{ij})_j$ and that almost all $e_i$ are mapped to zero. In particular, there is no homomorphism mapping all $e_i mapsto 1$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
answered Sep 26 '14 at 15:17
Martin BrandenburgMartin Brandenburg
108k13165335
108k13165335
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
add a comment |
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
where can I find the proof about there is no homomorphism mapping all e i ↦1? The original reference seems to be written in German..Could you suggest me some English links?
$endgroup$
– Keith
Sep 26 '14 at 17:12
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
I have included a link to MO.
$endgroup$
– Martin Brandenburg
Sep 26 '14 at 17:13
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
$begingroup$
Oh didn't recognize it. Thank you :)
$endgroup$
– Keith
Sep 26 '14 at 17:17
add a comment |
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$begingroup$
By $N$ you mean the set of natural numbers?
$endgroup$
– Daniel Fischer♦
Sep 26 '14 at 15:13