Can I Find A Map from a Module M to the kernel of a map p from M to M?
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I have a module homomorphism $p:Mrightarrow M$. I would like to find another module homomorphism $phi:Mrightarrow ker(p)$. Finding such a thing seems to be very challenging however. Is this possible? Note also that $p^2=p$.
modules
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
asked Nov 25 at 20:09
V. P. Sworski
11
add a comment |
up vote
0
down vote
favorite
I have a module homomorphism $p:Mrightarrow M$. I would like to find another module homomorphism $phi:Mrightarrow ker(p)$. Finding such a thing seems to be very challenging however. Is this possible? Note also that $p^2=p$.
modules
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
asked Nov 25 at 20:09
V. P. Sworski
11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a module homomorphism $p:Mrightarrow M$. I would like to find another module homomorphism $phi:Mrightarrow ker(p)$. Finding such a thing seems to be very challenging however. Is this possible? Note also that $p^2=p$.
modules
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
asked Nov 25 at 20:09
V. P. Sworski
11
I have a module homomorphism $p:Mrightarrow M$. I would like to find another module homomorphism $phi:Mrightarrow ker(p)$. Finding such a thing seems to be very challenging however. Is this possible? Note also that $p^2=p$.
modules
modules
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
asked Nov 25 at 20:09
V. P. Sworski
11
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
asked Nov 25 at 20:09
V. P. Sworski
11
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
edited Nov 25 at 20:16
Ashwin Trisal
1,1071515
1,1071515
asked Nov 25 at 20:09
V. P. Sworski
11
asked Nov 25 at 20:09
V. P. Sworski
11
asked Nov 25 at 20:09
V. P. Sworski
11
11
add a comment |
add a comment |
1 Answer
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Let $phi=1_M-p$. Then $pcirc phi=p(1_M-p)=p-p^2=p-p=0$, so $phi:Mtoker p$. But there are many such maps $phi$; taking $phi$ to be the $0$ map would also define a map from $M$ to $ker p$.
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $phi=1_M-p$. Then $pcirc phi=p(1_M-p)=p-p^2=p-p=0$, so $phi:Mtoker p$. But there are many such maps $phi$; taking $phi$ to be the $0$ map would also define a map from $M$ to $ker p$.
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
add a comment |
up vote
0
down vote
Let $phi=1_M-p$. Then $pcirc phi=p(1_M-p)=p-p^2=p-p=0$, so $phi:Mtoker p$. But there are many such maps $phi$; taking $phi$ to be the $0$ map would also define a map from $M$ to $ker p$.
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $phi=1_M-p$. Then $pcirc phi=p(1_M-p)=p-p^2=p-p=0$, so $phi:Mtoker p$. But there are many such maps $phi$; taking $phi$ to be the $0$ map would also define a map from $M$ to $ker p$.
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
Let $phi=1_M-p$. Then $pcirc phi=p(1_M-p)=p-p^2=p-p=0$, so $phi:Mtoker p$. But there are many such maps $phi$; taking $phi$ to be the $0$ map would also define a map from $M$ to $ker p$.
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
answered Nov 25 at 20:15
Ashwin Trisal
1,1071515
1,1071515
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
add a comment |
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
Ah, I see what I'm missing. I need the map to be surjective as well.
– V. P. Sworski
Nov 25 at 20:25
add a comment |
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