Mixing
Do we have $operatornameKerf^sharp =operatornameKerg^sharp$?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
1
Let $f: Yto X$ and $g:Zto X$ be two closed immersions of locally ringed spaces. If $Ysimeq Z$, do we have $operatornameKerf^sharp =operatornameKerg^sharp$?
algebraic-geometry sheaf-theory ringed-spaces
asked Jul 30 at 3:41
Born to be proud
41429
add a comment |Â
up vote
0
down vote
favorite
1
Let $f: Yto X$ and $g:Zto X$ be two closed immersions of locally ringed spaces. If $Ysimeq Z$, do we have $operatornameKerf^sharp =operatornameKerg^sharp$?
algebraic-geometry sheaf-theory ringed-spaces
asked Jul 30 at 3:41
Born to be proud
41429
An isomorphism of schemes $Ysimeq Z$ is not enough to conclude that $ker(f^sharp)=ker(g^sharp)$. If you have an isomorphism $k:Ysimeq Z$ such that $gcirc k=f$ (i.e. an isomorphism of $X$-schemes), then you can check (using the definition of the map of sheaves for a composition of maps of ringed spaces) that $ker(f^sharp)=ker(g^sharp)$.
– Keenan Kidwell
Jul 30 at 3:58
add a comment |Â
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
Let $f: Yto X$ and $g:Zto X$ be two closed immersions of locally ringed spaces. If $Ysimeq Z$, do we have $operatornameKerf^sharp =operatornameKerg^sharp$?
algebraic-geometry sheaf-theory ringed-spaces
asked Jul 30 at 3:41
Born to be proud
41429
Let $f: Yto X$ and $g:Zto X$ be two closed immersions of locally ringed spaces. If $Ysimeq Z$, do we have $operatornameKerf^sharp =operatornameKerg^sharp$?
algebraic-geometry sheaf-theory ringed-spaces
asked Jul 30 at 3:41
Born to be proud
41429
asked Jul 30 at 3:41
Born to be proud
41429
asked Jul 30 at 3:41
Born to be proud
41429
asked Jul 30 at 3:41
Born to be proud
41429
41429
An isomorphism of schemes $Ysimeq Z$ is not enough to conclude that $ker(f^sharp)=ker(g^sharp)$. If you have an isomorphism $k:Ysimeq Z$ such that $gcirc k=f$ (i.e. an isomorphism of $X$-schemes), then you can check (using the definition of the map of sheaves for a composition of maps of ringed spaces) that $ker(f^sharp)=ker(g^sharp)$.
– Keenan Kidwell
Jul 30 at 3:58
add a comment |Â
An isomorphism of schemes $Ysimeq Z$ is not enough to conclude that $ker(f^sharp)=ker(g^sharp)$. If you have an isomorphism $k:Ysimeq Z$ such that $gcirc k=f$ (i.e. an isomorphism of $X$-schemes), then you can check (using the definition of the map of sheaves for a composition of maps of ringed spaces) that $ker(f^sharp)=ker(g^sharp)$.
– Keenan Kidwell
Jul 30 at 3:58
An isomorphism of schemes $Ysimeq Z$ is not enough to conclude that $ker(f^sharp)=ker(g^sharp)$. If you have an isomorphism $k:Ysimeq Z$ such that $gcirc k=f$ (i.e. an isomorphism of $X$-schemes), then you can check (using the definition of the map of sheaves for a composition of maps of ringed spaces) that $ker(f^sharp)=ker(g^sharp)$.
– Keenan Kidwell
Jul 30 at 3:58
An isomorphism of schemes $Ysimeq Z$ is not enough to conclude that $ker(f^sharp)=ker(g^sharp)$. If you have an isomorphism $k:Ysimeq Z$ such that $gcirc k=f$ (i.e. an isomorphism of $X$-schemes), then you can check (using the definition of the map of sheaves for a composition of maps of ringed spaces) that $ker(f^sharp)=ker(g^sharp)$.
– Keenan Kidwell
Jul 30 at 3:58
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Â
draft saved
draft discarded
Â
draft saved
draft discarded
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%2f2866637%2fdo-we-have-operatornamekerf-sharp-operatornamekerg-sharp%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
An isomorphism of schemes $Ysimeq Z$ is not enough to conclude that $ker(f^sharp)=ker(g^sharp)$. If you have an isomorphism $k:Ysimeq Z$ such that $gcirc k=f$ (i.e. an isomorphism of $X$-schemes), then you can check (using the definition of the map of sheaves for a composition of maps of ringed spaces) that $ker(f^sharp)=ker(g^sharp)$.
– Keenan Kidwell
Jul 30 at 3:58