Surjective module homomorphism? $0$ module homomorphism?
$begingroup$
I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:
Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.
1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?
2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?
modules
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
$endgroup$
add a comment |
$begingroup$
I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:
Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.
1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?
2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?
modules
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
$endgroup$
1
$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30
$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31
add a comment |
$begingroup$
I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:
Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.
1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?
2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?
modules
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
$endgroup$
I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:
Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.
1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?
2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?
modules
modules
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
edited Dec 5 '18 at 19:35
zipirovich
11.3k11731
11.3k11731
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
asked Dec 5 '18 at 19:21
roi_saumonroi_saumon
57238
57238
1
$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30
$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31
add a comment |
1
$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30
$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31
1
1
$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30
$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30
$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31
$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31
add a comment |
2 Answers
2
active
oldest
votes
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For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.
For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
$endgroup$
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
add a comment |
$begingroup$
1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.
2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.
For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
$endgroup$
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
add a comment |
$begingroup$
For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.
For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
$endgroup$
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
add a comment |
$begingroup$
For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.
For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
$endgroup$
For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.
For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
edited Dec 5 '18 at 21:42
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
answered Dec 5 '18 at 19:33
ServaesServaes
26.8k34098
26.8k34098
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
add a comment |
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
$endgroup$
– roi_saumon
Dec 5 '18 at 21:27
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
$begingroup$
Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
$endgroup$
– Servaes
Dec 5 '18 at 21:42
add a comment |
$begingroup$
1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.
2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.
2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.
2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
$endgroup$
1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.
2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
answered Dec 5 '18 at 21:54
zipirovichzipirovich
11.3k11731
11.3k11731
add a comment |
add a comment |
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1
$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30
$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31