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If $f,g:Eto E$, then prove that $gcirc f=0 Longleftrightarrow operatornameIm fsubset ker g$ [duplicate]
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This question already has an answer here:
$g circ f$ trivial implies that $Im f subseteq ker g$ [closed]
3 answers
I have seen this kind of question answered here $g circ f$ trivial implies that $Im f subseteq ker g$, but it's not the same as mine. In my own case, I'm considering $f,g:Eto E$
Hence, I want to prove that the following are equivalent:
- $$gcirc f=0,$$
- $$operatornameIm fsubset ker g.$$
Can anyone help with a detailed proof or reference?
linear-algebra abstract-algebra
asked Jul 27 at 14:19
Mike
62112
marked as duplicate by Arnaud Mortier, amWhy, Shailesh, hardmath, José Carlos Santos linear-algebra
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Jul 27 at 16:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
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favorite
This question already has an answer here:
$g circ f$ trivial implies that $Im f subseteq ker g$ [closed]
3 answers
I have seen this kind of question answered here $g circ f$ trivial implies that $Im f subseteq ker g$, but it's not the same as mine. In my own case, I'm considering $f,g:Eto E$
Hence, I want to prove that the following are equivalent:
- $$gcirc f=0,$$
- $$operatornameIm fsubset ker g.$$
Can anyone help with a detailed proof or reference?
linear-algebra abstract-algebra
asked Jul 27 at 14:19
Mike
62112
marked as duplicate by Arnaud Mortier, amWhy, Shailesh, hardmath, José Carlos Santos linear-algebra
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Jul 27 at 16:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Isn't yours obtained from theirs by taking $X_1=X_2=X_3=E$?
– Calvin Khor
Jul 27 at 14:22
1
What exactly is causing you trouble? To cite Git Gud from the linked question, do you know the definition of $operatornameKer$ and $operatornameIm$?
– Arnaud Mortier
Jul 27 at 14:27
@ Arnaud Mortier: Yes, I know those definitions!
– Mike
Jul 27 at 15:38
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
This question already has an answer here:
$g circ f$ trivial implies that $Im f subseteq ker g$ [closed]
3 answers
I have seen this kind of question answered here $g circ f$ trivial implies that $Im f subseteq ker g$, but it's not the same as mine. In my own case, I'm considering $f,g:Eto E$
Hence, I want to prove that the following are equivalent:
- $$gcirc f=0,$$
- $$operatornameIm fsubset ker g.$$
Can anyone help with a detailed proof or reference?
linear-algebra abstract-algebra
asked Jul 27 at 14:19
Mike
62112
This question already has an answer here:
$g circ f$ trivial implies that $Im f subseteq ker g$ [closed]
3 answers
I have seen this kind of question answered here $g circ f$ trivial implies that $Im f subseteq ker g$, but it's not the same as mine. In my own case, I'm considering $f,g:Eto E$
Hence, I want to prove that the following are equivalent:
- $$gcirc f=0,$$
- $$operatornameIm fsubset ker g.$$
Can anyone help with a detailed proof or reference?
This question already has an answer here:
$g circ f$ trivial implies that $Im f subseteq ker g$ [closed]
3 answers
linear-algebra abstract-algebra
asked Jul 27 at 14:19
Mike
62112
edited Jul 27 at 15:37
asked Jul 27 at 14:19
Mike
62112
asked Jul 27 at 14:19
Mike
62112
asked Jul 27 at 14:19
Mike
62112
62112
marked as duplicate by Arnaud Mortier, amWhy, Shailesh, hardmath, José Carlos Santos linear-algebra
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Jul 27 at 16:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud Mortier, amWhy, Shailesh, hardmath, José Carlos Santos linear-algebra
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Jul 27 at 16:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Isn't yours obtained from theirs by taking $X_1=X_2=X_3=E$?
– Calvin Khor
Jul 27 at 14:22
1
What exactly is causing you trouble? To cite Git Gud from the linked question, do you know the definition of $operatornameKer$ and $operatornameIm$?
– Arnaud Mortier
Jul 27 at 14:27
@ Arnaud Mortier: Yes, I know those definitions!
– Mike
Jul 27 at 15:38
add a comment |Â
3
Isn't yours obtained from theirs by taking $X_1=X_2=X_3=E$?
– Calvin Khor
Jul 27 at 14:22
1
What exactly is causing you trouble? To cite Git Gud from the linked question, do you know the definition of $operatornameKer$ and $operatornameIm$?
– Arnaud Mortier
Jul 27 at 14:27
@ Arnaud Mortier: Yes, I know those definitions!
– Mike
Jul 27 at 15:38
3
3
Isn't yours obtained from theirs by taking $X_1=X_2=X_3=E$?
– Calvin Khor
Jul 27 at 14:22
Isn't yours obtained from theirs by taking $X_1=X_2=X_3=E$?
– Calvin Khor
Jul 27 at 14:22
1
1
What exactly is causing you trouble? To cite Git Gud from the linked question, do you know the definition of $operatornameKer$ and $operatornameIm$?
– Arnaud Mortier
Jul 27 at 14:27
What exactly is causing you trouble? To cite Git Gud from the linked question, do you know the definition of $operatornameKer$ and $operatornameIm$?
– Arnaud Mortier
Jul 27 at 14:27
@ Arnaud Mortier: Yes, I know those definitions!
– Mike
Jul 27 at 15:38
@ Arnaud Mortier: Yes, I know those definitions!
– Mike
Jul 27 at 15:38
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Here are some tips:
$(Rightarrow)$ start by assuming $gcirc f=0$. Then take any $xinoperatornameim(f)$ and try to show $xinker(g)$, which means you have to show $g(x)=0$. Hint: what does it mean for $x$ to be in $operatornameim(f)$?
$(Leftarrow)$ start by assuming $operatornameim(f)subseteqker(g)$. Then take any $xin E$ and show $(gcirc f)(x)=g(f(x))=0$. Notice $f(x)inoperatornameim(f)$.
answered Jul 27 at 14:31
Dave
7,8711831
add a comment |Â
up vote
1
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First $textim(f)=f(x)in Emid xin E$, the value of $f(x)$ as $x$ runs over all the elements in $E$. And $ker(g)=xin Emid g(x)=0$ is the set of elements $xin E$ such that $g(x)=0$
- If $gcirc f(x)=g(f(x))=0$, then $f(x)inker(g)$. But this holds for all values of $x$, so $textim(f)subseteqker(g)$.
Can you do the same for the other direction?
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
add a comment |Â
up vote
1
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$(Rightarrow)$ Assume that $gcirc f=0.$ We show that $operatornameIm fsubset ker g.$
$textLet;yin operatornameIm f,$ then $exists xin E$ such that $f(x)=y,$
$$g(f(x))=g(y)=0$$
$$implies g(f(x))=0$$
$$implies f(x)in ker g$$
$$implies yin ker g$$
$$implies operatornameIm fsubset ker g$$
$(Leftarrow)$ Assume that $operatornameIm fsubset ker g$. We show that $gcirc f=0.$
Let $xin E implies f(x)in operatornameIm f subset ker g$
$$implies f(x)in ker g$$
$$implies g(f(x))=0$$
$$implies gcirc f=0.$$
answered Jul 27 at 16:06
Mike
62112
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here are some tips:
$(Rightarrow)$ start by assuming $gcirc f=0$. Then take any $xinoperatornameim(f)$ and try to show $xinker(g)$, which means you have to show $g(x)=0$. Hint: what does it mean for $x$ to be in $operatornameim(f)$?
$(Leftarrow)$ start by assuming $operatornameim(f)subseteqker(g)$. Then take any $xin E$ and show $(gcirc f)(x)=g(f(x))=0$. Notice $f(x)inoperatornameim(f)$.
answered Jul 27 at 14:31
Dave
7,8711831
add a comment |Â
up vote
1
down vote
accepted
Here are some tips:
$(Rightarrow)$ start by assuming $gcirc f=0$. Then take any $xinoperatornameim(f)$ and try to show $xinker(g)$, which means you have to show $g(x)=0$. Hint: what does it mean for $x$ to be in $operatornameim(f)$?
$(Leftarrow)$ start by assuming $operatornameim(f)subseteqker(g)$. Then take any $xin E$ and show $(gcirc f)(x)=g(f(x))=0$. Notice $f(x)inoperatornameim(f)$.
answered Jul 27 at 14:31
Dave
7,8711831
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here are some tips:
$(Rightarrow)$ start by assuming $gcirc f=0$. Then take any $xinoperatornameim(f)$ and try to show $xinker(g)$, which means you have to show $g(x)=0$. Hint: what does it mean for $x$ to be in $operatornameim(f)$?
$(Leftarrow)$ start by assuming $operatornameim(f)subseteqker(g)$. Then take any $xin E$ and show $(gcirc f)(x)=g(f(x))=0$. Notice $f(x)inoperatornameim(f)$.
answered Jul 27 at 14:31
Dave
7,8711831
Here are some tips:
$(Rightarrow)$ start by assuming $gcirc f=0$. Then take any $xinoperatornameim(f)$ and try to show $xinker(g)$, which means you have to show $g(x)=0$. Hint: what does it mean for $x$ to be in $operatornameim(f)$?
$(Leftarrow)$ start by assuming $operatornameim(f)subseteqker(g)$. Then take any $xin E$ and show $(gcirc f)(x)=g(f(x))=0$. Notice $f(x)inoperatornameim(f)$.
answered Jul 27 at 14:31
Dave
7,8711831
answered Jul 27 at 14:31
Dave
7,8711831
answered Jul 27 at 14:31
Dave
7,8711831
answered Jul 27 at 14:31
Dave
7,8711831
7,8711831
add a comment |Â
add a comment |Â
up vote
1
down vote
First $textim(f)=f(x)in Emid xin E$, the value of $f(x)$ as $x$ runs over all the elements in $E$. And $ker(g)=xin Emid g(x)=0$ is the set of elements $xin E$ such that $g(x)=0$
- If $gcirc f(x)=g(f(x))=0$, then $f(x)inker(g)$. But this holds for all values of $x$, so $textim(f)subseteqker(g)$.
Can you do the same for the other direction?
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
add a comment |Â
up vote
1
down vote
First $textim(f)=f(x)in Emid xin E$, the value of $f(x)$ as $x$ runs over all the elements in $E$. And $ker(g)=xin Emid g(x)=0$ is the set of elements $xin E$ such that $g(x)=0$
- If $gcirc f(x)=g(f(x))=0$, then $f(x)inker(g)$. But this holds for all values of $x$, so $textim(f)subseteqker(g)$.
Can you do the same for the other direction?
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First $textim(f)=f(x)in Emid xin E$, the value of $f(x)$ as $x$ runs over all the elements in $E$. And $ker(g)=xin Emid g(x)=0$ is the set of elements $xin E$ such that $g(x)=0$
- If $gcirc f(x)=g(f(x))=0$, then $f(x)inker(g)$. But this holds for all values of $x$, so $textim(f)subseteqker(g)$.
Can you do the same for the other direction?
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
First $textim(f)=f(x)in Emid xin E$, the value of $f(x)$ as $x$ runs over all the elements in $E$. And $ker(g)=xin Emid g(x)=0$ is the set of elements $xin E$ such that $g(x)=0$
- If $gcirc f(x)=g(f(x))=0$, then $f(x)inker(g)$. But this holds for all values of $x$, so $textim(f)subseteqker(g)$.
Can you do the same for the other direction?
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
edited Jul 27 at 15:28
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
answered Jul 27 at 15:13
cansomeonehelpmeout
4,7673830
4,7673830
add a comment |Â
add a comment |Â
up vote
1
down vote
$(Rightarrow)$ Assume that $gcirc f=0.$ We show that $operatornameIm fsubset ker g.$
$textLet;yin operatornameIm f,$ then $exists xin E$ such that $f(x)=y,$
$$g(f(x))=g(y)=0$$
$$implies g(f(x))=0$$
$$implies f(x)in ker g$$
$$implies yin ker g$$
$$implies operatornameIm fsubset ker g$$
$(Leftarrow)$ Assume that $operatornameIm fsubset ker g$. We show that $gcirc f=0.$
Let $xin E implies f(x)in operatornameIm f subset ker g$
$$implies f(x)in ker g$$
$$implies g(f(x))=0$$
$$implies gcirc f=0.$$
answered Jul 27 at 16:06
Mike
62112
add a comment |Â
up vote
1
down vote
$(Rightarrow)$ Assume that $gcirc f=0.$ We show that $operatornameIm fsubset ker g.$
$textLet;yin operatornameIm f,$ then $exists xin E$ such that $f(x)=y,$
$$g(f(x))=g(y)=0$$
$$implies g(f(x))=0$$
$$implies f(x)in ker g$$
$$implies yin ker g$$
$$implies operatornameIm fsubset ker g$$
$(Leftarrow)$ Assume that $operatornameIm fsubset ker g$. We show that $gcirc f=0.$
Let $xin E implies f(x)in operatornameIm f subset ker g$
$$implies f(x)in ker g$$
$$implies g(f(x))=0$$
$$implies gcirc f=0.$$
answered Jul 27 at 16:06
Mike
62112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$(Rightarrow)$ Assume that $gcirc f=0.$ We show that $operatornameIm fsubset ker g.$
$textLet;yin operatornameIm f,$ then $exists xin E$ such that $f(x)=y,$
$$g(f(x))=g(y)=0$$
$$implies g(f(x))=0$$
$$implies f(x)in ker g$$
$$implies yin ker g$$
$$implies operatornameIm fsubset ker g$$
$(Leftarrow)$ Assume that $operatornameIm fsubset ker g$. We show that $gcirc f=0.$
Let $xin E implies f(x)in operatornameIm f subset ker g$
$$implies f(x)in ker g$$
$$implies g(f(x))=0$$
$$implies gcirc f=0.$$
answered Jul 27 at 16:06
Mike
62112
$(Rightarrow)$ Assume that $gcirc f=0.$ We show that $operatornameIm fsubset ker g.$
$textLet;yin operatornameIm f,$ then $exists xin E$ such that $f(x)=y,$
$$g(f(x))=g(y)=0$$
$$implies g(f(x))=0$$
$$implies f(x)in ker g$$
$$implies yin ker g$$
$$implies operatornameIm fsubset ker g$$
$(Leftarrow)$ Assume that $operatornameIm fsubset ker g$. We show that $gcirc f=0.$
Let $xin E implies f(x)in operatornameIm f subset ker g$
$$implies f(x)in ker g$$
$$implies g(f(x))=0$$
$$implies gcirc f=0.$$
answered Jul 27 at 16:06
Mike
62112
edited Jul 27 at 16:12
answered Jul 27 at 16:06
Mike
62112
answered Jul 27 at 16:06
Mike
62112
answered Jul 27 at 16:06
Mike
62112
62112
add a comment |Â
add a comment |Â
3
Isn't yours obtained from theirs by taking $X_1=X_2=X_3=E$?
– Calvin Khor
Jul 27 at 14:22
1
What exactly is causing you trouble? To cite Git Gud from the linked question, do you know the definition of $operatornameKer$ and $operatornameIm$?
– Arnaud Mortier
Jul 27 at 14:27
@ Arnaud Mortier: Yes, I know those definitions!
– Mike
Jul 27 at 15:38