Mixing
![How is $[P text AND (Q text OR R)] text IFF [(P text AND Q) text OR (P text AND R)]$ valid?](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Checking equality of kernel of module maps
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
This is a question on Cartan Eilenberg Homological Algebra Pg 42, Satellite.
Suppose given the following exact sequences of modules with $P$ projective over some $Lambda$ ring.
$0to A'to Rto Pto 0$
$0to A'to Ato A''to 0$
Further suppose $0to Mto Pto Ato 0$ and $0to Mto Rto Ato 0$ are exact s.t. the arrows induced $Pto A''$ and $Rto A$ form commutative diagram. This induces a unique arrow on $A'to A$. Assume $Mto M$ arrow is identity arrow and $Mto M,Mto P,Rto P$ and $Mto R$ forms a commutative square.
Let $T$ be a half exact additive functor.(i.e. Given $0to Ato Bto Cto 0$, then $TAto TBto TC$ is exact.)
Then $T$ acts on the diagram formed by previous 4 exact sequence yields the following exact sequences.
$0to T(M)to T(M)to 0$
$0to T(A')to T(R)to T(P)to 0$ is exact by $P$ projective and $T$ additive.($0$ of first exact sequence is aligned with $T(A')$ position.)
There will be arrows $0to T(A'),T(M)to T(R)$ and $T(M)to T(P)$. Then Snake lemma yields $Ker(T(M)to T(P))to T(A')to Coker(T(M)to T(R))$ is exact.
$textbfQ:$ The book says "$T(M)to T(R)to T(A)$ is exact, it follows $Ker(T(A')to Coker(T(M)to T(R)))=Ker(T(A')to T(A))$." How is this supposed to be obvious? I did verify the equality but I do not see any particular reason this is obvious. Or was kernel universal property invoked here?
abstract-algebra homological-algebra
asked Jul 16 at 22:33
user45765
2,1942718
add a comment |Â
up vote
1
down vote
favorite
This is a question on Cartan Eilenberg Homological Algebra Pg 42, Satellite.
Suppose given the following exact sequences of modules with $P$ projective over some $Lambda$ ring.
$0to A'to Rto Pto 0$
$0to A'to Ato A''to 0$
Further suppose $0to Mto Pto Ato 0$ and $0to Mto Rto Ato 0$ are exact s.t. the arrows induced $Pto A''$ and $Rto A$ form commutative diagram. This induces a unique arrow on $A'to A$. Assume $Mto M$ arrow is identity arrow and $Mto M,Mto P,Rto P$ and $Mto R$ forms a commutative square.
Let $T$ be a half exact additive functor.(i.e. Given $0to Ato Bto Cto 0$, then $TAto TBto TC$ is exact.)
Then $T$ acts on the diagram formed by previous 4 exact sequence yields the following exact sequences.
$0to T(M)to T(M)to 0$
$0to T(A')to T(R)to T(P)to 0$ is exact by $P$ projective and $T$ additive.($0$ of first exact sequence is aligned with $T(A')$ position.)
There will be arrows $0to T(A'),T(M)to T(R)$ and $T(M)to T(P)$. Then Snake lemma yields $Ker(T(M)to T(P))to T(A')to Coker(T(M)to T(R))$ is exact.
$textbfQ:$ The book says "$T(M)to T(R)to T(A)$ is exact, it follows $Ker(T(A')to Coker(T(M)to T(R)))=Ker(T(A')to T(A))$." How is this supposed to be obvious? I did verify the equality but I do not see any particular reason this is obvious. Or was kernel universal property invoked here?
abstract-algebra homological-algebra
asked Jul 16 at 22:33
user45765
2,1942718
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is a question on Cartan Eilenberg Homological Algebra Pg 42, Satellite.
Suppose given the following exact sequences of modules with $P$ projective over some $Lambda$ ring.
$0to A'to Rto Pto 0$
$0to A'to Ato A''to 0$
Further suppose $0to Mto Pto Ato 0$ and $0to Mto Rto Ato 0$ are exact s.t. the arrows induced $Pto A''$ and $Rto A$ form commutative diagram. This induces a unique arrow on $A'to A$. Assume $Mto M$ arrow is identity arrow and $Mto M,Mto P,Rto P$ and $Mto R$ forms a commutative square.
Let $T$ be a half exact additive functor.(i.e. Given $0to Ato Bto Cto 0$, then $TAto TBto TC$ is exact.)
Then $T$ acts on the diagram formed by previous 4 exact sequence yields the following exact sequences.
$0to T(M)to T(M)to 0$
$0to T(A')to T(R)to T(P)to 0$ is exact by $P$ projective and $T$ additive.($0$ of first exact sequence is aligned with $T(A')$ position.)
There will be arrows $0to T(A'),T(M)to T(R)$ and $T(M)to T(P)$. Then Snake lemma yields $Ker(T(M)to T(P))to T(A')to Coker(T(M)to T(R))$ is exact.
$textbfQ:$ The book says "$T(M)to T(R)to T(A)$ is exact, it follows $Ker(T(A')to Coker(T(M)to T(R)))=Ker(T(A')to T(A))$." How is this supposed to be obvious? I did verify the equality but I do not see any particular reason this is obvious. Or was kernel universal property invoked here?
abstract-algebra homological-algebra
asked Jul 16 at 22:33
user45765
2,1942718
This is a question on Cartan Eilenberg Homological Algebra Pg 42, Satellite.
Suppose given the following exact sequences of modules with $P$ projective over some $Lambda$ ring.
$0to A'to Rto Pto 0$
$0to A'to Ato A''to 0$
Further suppose $0to Mto Pto Ato 0$ and $0to Mto Rto Ato 0$ are exact s.t. the arrows induced $Pto A''$ and $Rto A$ form commutative diagram. This induces a unique arrow on $A'to A$. Assume $Mto M$ arrow is identity arrow and $Mto M,Mto P,Rto P$ and $Mto R$ forms a commutative square.
Let $T$ be a half exact additive functor.(i.e. Given $0to Ato Bto Cto 0$, then $TAto TBto TC$ is exact.)
Then $T$ acts on the diagram formed by previous 4 exact sequence yields the following exact sequences.
$0to T(M)to T(M)to 0$
$0to T(A')to T(R)to T(P)to 0$ is exact by $P$ projective and $T$ additive.($0$ of first exact sequence is aligned with $T(A')$ position.)
There will be arrows $0to T(A'),T(M)to T(R)$ and $T(M)to T(P)$. Then Snake lemma yields $Ker(T(M)to T(P))to T(A')to Coker(T(M)to T(R))$ is exact.
$textbfQ:$ The book says "$T(M)to T(R)to T(A)$ is exact, it follows $Ker(T(A')to Coker(T(M)to T(R)))=Ker(T(A')to T(A))$." How is this supposed to be obvious? I did verify the equality but I do not see any particular reason this is obvious. Or was kernel universal property invoked here?
abstract-algebra homological-algebra
asked Jul 16 at 22:33
user45765
2,1942718
asked Jul 16 at 22:33
user45765
2,1942718
asked Jul 16 at 22:33
user45765
2,1942718
asked Jul 16 at 22:33
user45765
2,1942718
2,1942718
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I think the following is the most direct reasoning, although perhaps you already reasoned this way:
As $T(M)xrightarrowf T(R)xrightarrowg T(A)$ is exact, $operatornamecoker(f)rightarrow T(A)$ is monic; thus for any $Xrightarrow T(R)$, we have $ker(Xrightarrow operatornamecoker(f)) = ker(Xrightarrow T(A))$.
answered Jul 17 at 0:21
ne-
1067
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think the following is the most direct reasoning, although perhaps you already reasoned this way:
As $T(M)xrightarrowf T(R)xrightarrowg T(A)$ is exact, $operatornamecoker(f)rightarrow T(A)$ is monic; thus for any $Xrightarrow T(R)$, we have $ker(Xrightarrow operatornamecoker(f)) = ker(Xrightarrow T(A))$.
answered Jul 17 at 0:21
ne-
1067
add a comment |Â
up vote
1
down vote
accepted
I think the following is the most direct reasoning, although perhaps you already reasoned this way:
As $T(M)xrightarrowf T(R)xrightarrowg T(A)$ is exact, $operatornamecoker(f)rightarrow T(A)$ is monic; thus for any $Xrightarrow T(R)$, we have $ker(Xrightarrow operatornamecoker(f)) = ker(Xrightarrow T(A))$.
answered Jul 17 at 0:21
ne-
1067
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think the following is the most direct reasoning, although perhaps you already reasoned this way:
As $T(M)xrightarrowf T(R)xrightarrowg T(A)$ is exact, $operatornamecoker(f)rightarrow T(A)$ is monic; thus for any $Xrightarrow T(R)$, we have $ker(Xrightarrow operatornamecoker(f)) = ker(Xrightarrow T(A))$.
answered Jul 17 at 0:21
ne-
1067
I think the following is the most direct reasoning, although perhaps you already reasoned this way:
As $T(M)xrightarrowf T(R)xrightarrowg T(A)$ is exact, $operatornamecoker(f)rightarrow T(A)$ is monic; thus for any $Xrightarrow T(R)$, we have $ker(Xrightarrow operatornamecoker(f)) = ker(Xrightarrow T(A))$.
answered Jul 17 at 0:21
ne-
1067
answered Jul 17 at 0:21
ne-
1067
answered Jul 17 at 0:21
ne-
1067
answered Jul 17 at 0:21
ne-
1067
1067
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853926%2fchecking-equality-of-kernel-of-module-maps%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password