Mixing
The images of the identity maps under the connecting homomorphisms
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$newcommandExtoperatornameExt^1$I am studying the Hilton & Rees article on natural transformations between $Ext$ functors which can be found here.
More specifically, in Theorem 1.3 they prove that every natural transformation $varphi:Ext(B,-)to Ext(A,-)$ is induced by a homomorphism $etain hom(A,B)$.
They use the following result:
Let $requireAMScdbeginCD
0 @>>>A^primeprime @>>> A^prime @>>> A @>>> 0
endCD$ be a short exact sequence of modules.
We have connecting homomorphisms $requireAMScdbeginCD
ldots @>>> hom(A^primeprime,A^primeprime) @>partial>> Ext(A,A^primeprime) @>>>ldots
endCD$
and
$requireAMScdbeginCD
ldots @>>>hom(A,A) @>delta>> Ext(A,A^primeprime) @>>>ldots
endCD$
Then $partial(id_A^primeprime)+delta(id_A)=0$
I can't prove this result. The article references Cartier's article which can be found here. I had a very hard time studying the latter. I couldn't find the proof and my French is really weak.
Any help with the proof of this result would be greatly appreciated.
Thank you.
homological-algebra functors
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 12:23
Nick Papadopoulos
351213
add a comment |Â
up vote
1
down vote
favorite
$newcommandExtoperatornameExt^1$I am studying the Hilton & Rees article on natural transformations between $Ext$ functors which can be found here.
More specifically, in Theorem 1.3 they prove that every natural transformation $varphi:Ext(B,-)to Ext(A,-)$ is induced by a homomorphism $etain hom(A,B)$.
They use the following result:
Let $requireAMScdbeginCD
0 @>>>A^primeprime @>>> A^prime @>>> A @>>> 0
endCD$ be a short exact sequence of modules.
We have connecting homomorphisms $requireAMScdbeginCD
ldots @>>> hom(A^primeprime,A^primeprime) @>partial>> Ext(A,A^primeprime) @>>>ldots
endCD$
and
$requireAMScdbeginCD
ldots @>>>hom(A,A) @>delta>> Ext(A,A^primeprime) @>>>ldots
endCD$
Then $partial(id_A^primeprime)+delta(id_A)=0$
I can't prove this result. The article references Cartier's article which can be found here. I had a very hard time studying the latter. I couldn't find the proof and my French is really weak.
Any help with the proof of this result would be greatly appreciated.
Thank you.
homological-algebra functors
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 12:23
Nick Papadopoulos
351213
I cannot access the article by Hilton and Rees, so I cannot tell where they are pointing you to (the obvious guess is the last section), but if you tell me what's troubling you there I can help with figuring out what Cartier is saying.
– Pedro Tamaroff♦
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$newcommandExtoperatornameExt^1$I am studying the Hilton & Rees article on natural transformations between $Ext$ functors which can be found here.
More specifically, in Theorem 1.3 they prove that every natural transformation $varphi:Ext(B,-)to Ext(A,-)$ is induced by a homomorphism $etain hom(A,B)$.
They use the following result:
Let $requireAMScdbeginCD
0 @>>>A^primeprime @>>> A^prime @>>> A @>>> 0
endCD$ be a short exact sequence of modules.
We have connecting homomorphisms $requireAMScdbeginCD
ldots @>>> hom(A^primeprime,A^primeprime) @>partial>> Ext(A,A^primeprime) @>>>ldots
endCD$
and
$requireAMScdbeginCD
ldots @>>>hom(A,A) @>delta>> Ext(A,A^primeprime) @>>>ldots
endCD$
Then $partial(id_A^primeprime)+delta(id_A)=0$
I can't prove this result. The article references Cartier's article which can be found here. I had a very hard time studying the latter. I couldn't find the proof and my French is really weak.
Any help with the proof of this result would be greatly appreciated.
Thank you.
homological-algebra functors
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 12:23
Nick Papadopoulos
351213
$newcommandExtoperatornameExt^1$I am studying the Hilton & Rees article on natural transformations between $Ext$ functors which can be found here.
More specifically, in Theorem 1.3 they prove that every natural transformation $varphi:Ext(B,-)to Ext(A,-)$ is induced by a homomorphism $etain hom(A,B)$.
They use the following result:
Let $requireAMScdbeginCD
0 @>>>A^primeprime @>>> A^prime @>>> A @>>> 0
endCD$ be a short exact sequence of modules.
We have connecting homomorphisms $requireAMScdbeginCD
ldots @>>> hom(A^primeprime,A^primeprime) @>partial>> Ext(A,A^primeprime) @>>>ldots
endCD$
and
$requireAMScdbeginCD
ldots @>>>hom(A,A) @>delta>> Ext(A,A^primeprime) @>>>ldots
endCD$
Then $partial(id_A^primeprime)+delta(id_A)=0$
I can't prove this result. The article references Cartier's article which can be found here. I had a very hard time studying the latter. I couldn't find the proof and my French is really weak.
Any help with the proof of this result would be greatly appreciated.
Thank you.
homological-algebra functors
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
asked Aug 3 at 12:23
Nick Papadopoulos
351213
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
edited 2 days ago
Pedro Tamaroff♦
93.6k10143290
93.6k10143290
asked Aug 3 at 12:23
Nick Papadopoulos
351213
asked Aug 3 at 12:23
Nick Papadopoulos
351213
asked Aug 3 at 12:23
Nick Papadopoulos
351213
351213
I cannot access the article by Hilton and Rees, so I cannot tell where they are pointing you to (the obvious guess is the last section), but if you tell me what's troubling you there I can help with figuring out what Cartier is saying.
– Pedro Tamaroff♦
2 days ago
add a comment |Â
I cannot access the article by Hilton and Rees, so I cannot tell where they are pointing you to (the obvious guess is the last section), but if you tell me what's troubling you there I can help with figuring out what Cartier is saying.
– Pedro Tamaroff♦
2 days ago
I cannot access the article by Hilton and Rees, so I cannot tell where they are pointing you to (the obvious guess is the last section), but if you tell me what's troubling you there I can help with figuring out what Cartier is saying.
– Pedro Tamaroff♦
2 days ago
I cannot access the article by Hilton and Rees, so I cannot tell where they are pointing you to (the obvious guess is the last section), but if you tell me what's troubling you there I can help with figuring out what Cartier is saying.
– Pedro Tamaroff♦
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Add If you have set up the machinery of suspended (or triangulated) categories, it seems the proof is the following: your SEC is a morphism $xi : A'to A''[1]$, and $partial(1_A'') = 1_A''[1]circ xi = -xi$ since translating the identity introduces a sign, while $delta(1_A') = xi circ 1_A' = xi$.
$newcommandExtoperatornameExt^1$Let me write your exact sequence $S$ and fix another module $B$, so that you want to describe two connecting morphisms
$$partial:hom_R(A'',B)to Ext_R(A',B),
qquaddelta:hom_R(B,A')to Ext_R(A',A'').$$
So take your SEC $0to A''to Ato A'to 0$. You want to describe the following connecting maps somehow and show that $partial 1_A''+delta 1_A'=0$. One way to proceed is as follows. Take a map $f:A''to B$, and consider the diagram
beginCD
0 @>>>B @>>> E @>>> A'@>>>0\
&&@AfAA @AAA @AA1_A'A \
0@>>>A'' @>>> A @>>> A'@>>>0 \
endCD
where the first square is a pushout. Then the class of the extension in the first row is $partial f$. Similarly, if you take $g:Bto A'$ and consider the diagram where the second square is a pullback, then the class of the extension in the second row is $delta g$.
beginCD
0 @>>>A'' @>>> A @>>> A'@>>>0\
&&@A1_A''AA @AAA @AAgA \
0@>>>A'' @>>> F @>>> A'@>>>0 \
endCD
We now consider your case, where the SEC $S$ goes (almost) to itself. I claim that the extension $partial 1_A''$ is isomorphic to the extension
$$0to A''to Ato A'to 0$$
where the first arrow, say $f$, now has a minus sign, that is, it is $-f$.
Indeed, when you consider the pushout, $E = A''oplus A /langle (a'',-fa'')rangle$ and the induced map $A''to E simeq A$ is given by
$$a''to textrmclass(a'',0) = textrmclass(0,-fa'') simeq -fa''in A,$$
while the projection $Eto E/langle (0,-fa'')rangle = operatornamecoker f simeq A$ identifies with $g$, without signs.
On the other hand, when doing the pullback, no signs are involved. To conclude, the sum of two extensions represented by SECs is given by the Baer sum, and it is a fact that negating one of the morphisms in a SEC gives you its additive inverse for the Baer sum.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Add If you have set up the machinery of suspended (or triangulated) categories, it seems the proof is the following: your SEC is a morphism $xi : A'to A''[1]$, and $partial(1_A'') = 1_A''[1]circ xi = -xi$ since translating the identity introduces a sign, while $delta(1_A') = xi circ 1_A' = xi$.
$newcommandExtoperatornameExt^1$Let me write your exact sequence $S$ and fix another module $B$, so that you want to describe two connecting morphisms
$$partial:hom_R(A'',B)to Ext_R(A',B),
qquaddelta:hom_R(B,A')to Ext_R(A',A'').$$
So take your SEC $0to A''to Ato A'to 0$. You want to describe the following connecting maps somehow and show that $partial 1_A''+delta 1_A'=0$. One way to proceed is as follows. Take a map $f:A''to B$, and consider the diagram
beginCD
0 @>>>B @>>> E @>>> A'@>>>0\
&&@AfAA @AAA @AA1_A'A \
0@>>>A'' @>>> A @>>> A'@>>>0 \
endCD
where the first square is a pushout. Then the class of the extension in the first row is $partial f$. Similarly, if you take $g:Bto A'$ and consider the diagram where the second square is a pullback, then the class of the extension in the second row is $delta g$.
beginCD
0 @>>>A'' @>>> A @>>> A'@>>>0\
&&@A1_A''AA @AAA @AAgA \
0@>>>A'' @>>> F @>>> A'@>>>0 \
endCD
We now consider your case, where the SEC $S$ goes (almost) to itself. I claim that the extension $partial 1_A''$ is isomorphic to the extension
$$0to A''to Ato A'to 0$$
where the first arrow, say $f$, now has a minus sign, that is, it is $-f$.
Indeed, when you consider the pushout, $E = A''oplus A /langle (a'',-fa'')rangle$ and the induced map $A''to E simeq A$ is given by
$$a''to textrmclass(a'',0) = textrmclass(0,-fa'') simeq -fa''in A,$$
while the projection $Eto E/langle (0,-fa'')rangle = operatornamecoker f simeq A$ identifies with $g$, without signs.
On the other hand, when doing the pullback, no signs are involved. To conclude, the sum of two extensions represented by SECs is given by the Baer sum, and it is a fact that negating one of the morphisms in a SEC gives you its additive inverse for the Baer sum.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
0
down vote
Add If you have set up the machinery of suspended (or triangulated) categories, it seems the proof is the following: your SEC is a morphism $xi : A'to A''[1]$, and $partial(1_A'') = 1_A''[1]circ xi = -xi$ since translating the identity introduces a sign, while $delta(1_A') = xi circ 1_A' = xi$.
$newcommandExtoperatornameExt^1$Let me write your exact sequence $S$ and fix another module $B$, so that you want to describe two connecting morphisms
$$partial:hom_R(A'',B)to Ext_R(A',B),
qquaddelta:hom_R(B,A')to Ext_R(A',A'').$$
So take your SEC $0to A''to Ato A'to 0$. You want to describe the following connecting maps somehow and show that $partial 1_A''+delta 1_A'=0$. One way to proceed is as follows. Take a map $f:A''to B$, and consider the diagram
beginCD
0 @>>>B @>>> E @>>> A'@>>>0\
&&@AfAA @AAA @AA1_A'A \
0@>>>A'' @>>> A @>>> A'@>>>0 \
endCD
where the first square is a pushout. Then the class of the extension in the first row is $partial f$. Similarly, if you take $g:Bto A'$ and consider the diagram where the second square is a pullback, then the class of the extension in the second row is $delta g$.
beginCD
0 @>>>A'' @>>> A @>>> A'@>>>0\
&&@A1_A''AA @AAA @AAgA \
0@>>>A'' @>>> F @>>> A'@>>>0 \
endCD
We now consider your case, where the SEC $S$ goes (almost) to itself. I claim that the extension $partial 1_A''$ is isomorphic to the extension
$$0to A''to Ato A'to 0$$
where the first arrow, say $f$, now has a minus sign, that is, it is $-f$.
Indeed, when you consider the pushout, $E = A''oplus A /langle (a'',-fa'')rangle$ and the induced map $A''to E simeq A$ is given by
$$a''to textrmclass(a'',0) = textrmclass(0,-fa'') simeq -fa''in A,$$
while the projection $Eto E/langle (0,-fa'')rangle = operatornamecoker f simeq A$ identifies with $g$, without signs.
On the other hand, when doing the pullback, no signs are involved. To conclude, the sum of two extensions represented by SECs is given by the Baer sum, and it is a fact that negating one of the morphisms in a SEC gives you its additive inverse for the Baer sum.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Add If you have set up the machinery of suspended (or triangulated) categories, it seems the proof is the following: your SEC is a morphism $xi : A'to A''[1]$, and $partial(1_A'') = 1_A''[1]circ xi = -xi$ since translating the identity introduces a sign, while $delta(1_A') = xi circ 1_A' = xi$.
$newcommandExtoperatornameExt^1$Let me write your exact sequence $S$ and fix another module $B$, so that you want to describe two connecting morphisms
$$partial:hom_R(A'',B)to Ext_R(A',B),
qquaddelta:hom_R(B,A')to Ext_R(A',A'').$$
So take your SEC $0to A''to Ato A'to 0$. You want to describe the following connecting maps somehow and show that $partial 1_A''+delta 1_A'=0$. One way to proceed is as follows. Take a map $f:A''to B$, and consider the diagram
beginCD
0 @>>>B @>>> E @>>> A'@>>>0\
&&@AfAA @AAA @AA1_A'A \
0@>>>A'' @>>> A @>>> A'@>>>0 \
endCD
where the first square is a pushout. Then the class of the extension in the first row is $partial f$. Similarly, if you take $g:Bto A'$ and consider the diagram where the second square is a pullback, then the class of the extension in the second row is $delta g$.
beginCD
0 @>>>A'' @>>> A @>>> A'@>>>0\
&&@A1_A''AA @AAA @AAgA \
0@>>>A'' @>>> F @>>> A'@>>>0 \
endCD
We now consider your case, where the SEC $S$ goes (almost) to itself. I claim that the extension $partial 1_A''$ is isomorphic to the extension
$$0to A''to Ato A'to 0$$
where the first arrow, say $f$, now has a minus sign, that is, it is $-f$.
Indeed, when you consider the pushout, $E = A''oplus A /langle (a'',-fa'')rangle$ and the induced map $A''to E simeq A$ is given by
$$a''to textrmclass(a'',0) = textrmclass(0,-fa'') simeq -fa''in A,$$
while the projection $Eto E/langle (0,-fa'')rangle = operatornamecoker f simeq A$ identifies with $g$, without signs.
On the other hand, when doing the pullback, no signs are involved. To conclude, the sum of two extensions represented by SECs is given by the Baer sum, and it is a fact that negating one of the morphisms in a SEC gives you its additive inverse for the Baer sum.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
Add If you have set up the machinery of suspended (or triangulated) categories, it seems the proof is the following: your SEC is a morphism $xi : A'to A''[1]$, and $partial(1_A'') = 1_A''[1]circ xi = -xi$ since translating the identity introduces a sign, while $delta(1_A') = xi circ 1_A' = xi$.
$newcommandExtoperatornameExt^1$Let me write your exact sequence $S$ and fix another module $B$, so that you want to describe two connecting morphisms
$$partial:hom_R(A'',B)to Ext_R(A',B),
qquaddelta:hom_R(B,A')to Ext_R(A',A'').$$
So take your SEC $0to A''to Ato A'to 0$. You want to describe the following connecting maps somehow and show that $partial 1_A''+delta 1_A'=0$. One way to proceed is as follows. Take a map $f:A''to B$, and consider the diagram
beginCD
0 @>>>B @>>> E @>>> A'@>>>0\
&&@AfAA @AAA @AA1_A'A \
0@>>>A'' @>>> A @>>> A'@>>>0 \
endCD
where the first square is a pushout. Then the class of the extension in the first row is $partial f$. Similarly, if you take $g:Bto A'$ and consider the diagram where the second square is a pullback, then the class of the extension in the second row is $delta g$.
beginCD
0 @>>>A'' @>>> A @>>> A'@>>>0\
&&@A1_A''AA @AAA @AAgA \
0@>>>A'' @>>> F @>>> A'@>>>0 \
endCD
We now consider your case, where the SEC $S$ goes (almost) to itself. I claim that the extension $partial 1_A''$ is isomorphic to the extension
$$0to A''to Ato A'to 0$$
where the first arrow, say $f$, now has a minus sign, that is, it is $-f$.
Indeed, when you consider the pushout, $E = A''oplus A /langle (a'',-fa'')rangle$ and the induced map $A''to E simeq A$ is given by
$$a''to textrmclass(a'',0) = textrmclass(0,-fa'') simeq -fa''in A,$$
while the projection $Eto E/langle (0,-fa'')rangle = operatornamecoker f simeq A$ identifies with $g$, without signs.
On the other hand, when doing the pullback, no signs are involved. To conclude, the sum of two extensions represented by SECs is given by the Baer sum, and it is a fact that negating one of the morphisms in a SEC gives you its additive inverse for the Baer sum.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
edited 2 days ago
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
93.6k10143290
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%2f2871014%2fthe-images-of-the-identity-maps-under-the-connecting-homomorphisms%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
I cannot access the article by Hilton and Rees, so I cannot tell where they are pointing you to (the obvious guess is the last section), but if you tell me what's troubling you there I can help with figuring out what Cartier is saying.
– Pedro Tamaroff♦
2 days ago