Mixing
Hartshorne Exercise II. 1.18
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
Exercise 1.18 of chapter II asks to show that given a map of topological spaces $f:Xto Y$, the functors $f_* : mathsfSh_Y longrightarrowmathsfSh_X$ and $ f^-1: mathsfSh_Xlongrightarrow mathsfSh_Y$ are adjoint. The suggestion is to define the unit and the counit and then check they satisfy the usual equations. Now I have defined the unit and counit, and want to check that if $mathscr G$ is a sheaf on $Y$ then
$$varepsilon_f^-1mathscr Gcirc f^-1(eta_mathscr G) : f^-1mathscr Gto f^-1f_*f^-1mathscr Gto f^-1mathscr G$$
is the identity, and similarly for sheaves on $X$. This is a bit cumbersome, and the only idea I have to bypass this is perhaps to look at the maps on stalks, which I think are easily seen to be the identity because of how $f_*$, $f^-1$ and the unit and counit are defined. Does this suffice? If not, what is the "right" way to proceed here?
algebraic-geometry sheaf-theory adjoint-functors
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
5
down vote
favorite
Exercise 1.18 of chapter II asks to show that given a map of topological spaces $f:Xto Y$, the functors $f_* : mathsfSh_Y longrightarrowmathsfSh_X$ and $ f^-1: mathsfSh_Xlongrightarrow mathsfSh_Y$ are adjoint. The suggestion is to define the unit and the counit and then check they satisfy the usual equations. Now I have defined the unit and counit, and want to check that if $mathscr G$ is a sheaf on $Y$ then
$$varepsilon_f^-1mathscr Gcirc f^-1(eta_mathscr G) : f^-1mathscr Gto f^-1f_*f^-1mathscr Gto f^-1mathscr G$$
is the identity, and similarly for sheaves on $X$. This is a bit cumbersome, and the only idea I have to bypass this is perhaps to look at the maps on stalks, which I think are easily seen to be the identity because of how $f_*$, $f^-1$ and the unit and counit are defined. Does this suffice? If not, what is the "right" way to proceed here?
algebraic-geometry sheaf-theory adjoint-functors
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
If it's the sheafification part of the definition of $f^-1$ that's tripping you up, you might try first proving $f_* : PSh_Y to PSh_X$ and $f^circ : PSh_X to PSh_Y$ are adjoint; then use that sheafification is also a left adjoint of the inclusion $Sh_X to PSh_X$.
– Daniel Schepler
Jul 31 at 21:35
@DanielSchepler Ah, yes, that came to mind but I didn't went further with it. Thanks.
– Pedro Tamaroff♦
Jul 31 at 21:42
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Exercise 1.18 of chapter II asks to show that given a map of topological spaces $f:Xto Y$, the functors $f_* : mathsfSh_Y longrightarrowmathsfSh_X$ and $ f^-1: mathsfSh_Xlongrightarrow mathsfSh_Y$ are adjoint. The suggestion is to define the unit and the counit and then check they satisfy the usual equations. Now I have defined the unit and counit, and want to check that if $mathscr G$ is a sheaf on $Y$ then
$$varepsilon_f^-1mathscr Gcirc f^-1(eta_mathscr G) : f^-1mathscr Gto f^-1f_*f^-1mathscr Gto f^-1mathscr G$$
is the identity, and similarly for sheaves on $X$. This is a bit cumbersome, and the only idea I have to bypass this is perhaps to look at the maps on stalks, which I think are easily seen to be the identity because of how $f_*$, $f^-1$ and the unit and counit are defined. Does this suffice? If not, what is the "right" way to proceed here?
algebraic-geometry sheaf-theory adjoint-functors
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
Exercise 1.18 of chapter II asks to show that given a map of topological spaces $f:Xto Y$, the functors $f_* : mathsfSh_Y longrightarrowmathsfSh_X$ and $ f^-1: mathsfSh_Xlongrightarrow mathsfSh_Y$ are adjoint. The suggestion is to define the unit and the counit and then check they satisfy the usual equations. Now I have defined the unit and counit, and want to check that if $mathscr G$ is a sheaf on $Y$ then
$$varepsilon_f^-1mathscr Gcirc f^-1(eta_mathscr G) : f^-1mathscr Gto f^-1f_*f^-1mathscr Gto f^-1mathscr G$$
is the identity, and similarly for sheaves on $X$. This is a bit cumbersome, and the only idea I have to bypass this is perhaps to look at the maps on stalks, which I think are easily seen to be the identity because of how $f_*$, $f^-1$ and the unit and counit are defined. Does this suffice? If not, what is the "right" way to proceed here?
algebraic-geometry sheaf-theory adjoint-functors
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
edited Jul 31 at 21:27
Eric Wofsey
161k12188297
161k12188297
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
asked Jul 31 at 21:09
Pedro Tamaroff♦
93.6k10143290
93.6k10143290
If it's the sheafification part of the definition of $f^-1$ that's tripping you up, you might try first proving $f_* : PSh_Y to PSh_X$ and $f^circ : PSh_X to PSh_Y$ are adjoint; then use that sheafification is also a left adjoint of the inclusion $Sh_X to PSh_X$.
– Daniel Schepler
Jul 31 at 21:35
@DanielSchepler Ah, yes, that came to mind but I didn't went further with it. Thanks.
– Pedro Tamaroff♦
Jul 31 at 21:42
add a comment |Â
If it's the sheafification part of the definition of $f^-1$ that's tripping you up, you might try first proving $f_* : PSh_Y to PSh_X$ and $f^circ : PSh_X to PSh_Y$ are adjoint; then use that sheafification is also a left adjoint of the inclusion $Sh_X to PSh_X$.
– Daniel Schepler
Jul 31 at 21:35
@DanielSchepler Ah, yes, that came to mind but I didn't went further with it. Thanks.
– Pedro Tamaroff♦
Jul 31 at 21:42
If it's the sheafification part of the definition of $f^-1$ that's tripping you up, you might try first proving $f_* : PSh_Y to PSh_X$ and $f^circ : PSh_X to PSh_Y$ are adjoint; then use that sheafification is also a left adjoint of the inclusion $Sh_X to PSh_X$.
– Daniel Schepler
Jul 31 at 21:35
If it's the sheafification part of the definition of $f^-1$ that's tripping you up, you might try first proving $f_* : PSh_Y to PSh_X$ and $f^circ : PSh_X to PSh_Y$ are adjoint; then use that sheafification is also a left adjoint of the inclusion $Sh_X to PSh_X$.
– Daniel Schepler
Jul 31 at 21:35
@DanielSchepler Ah, yes, that came to mind but I didn't went further with it. Thanks.
– Pedro Tamaroff♦
Jul 31 at 21:42
@DanielSchepler Ah, yes, that came to mind but I didn't went further with it. Thanks.
– Pedro Tamaroff♦
Jul 31 at 21:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:Fto G$ are morphisms which are not equal. Then there is some open $Usubseteq X$ and some $sin F(U)$ such that $a(s)neq b(s)$. Since a section of a sheaf is determined by its germs in each stalk, $a(s)neq b(s)$ means there exists some $xin U$ such that $a(s)_xneq b(s)_x$ in $G_x$. But $a(s)_x=a_x(s_x)$ and $b(s)_s=b_x(s_x)$, so this proves that $a_xneq b_x$. That is, there is some stalk on which $a$ and $b$ are not equal.
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:Fto G$ are morphisms which are not equal. Then there is some open $Usubseteq X$ and some $sin F(U)$ such that $a(s)neq b(s)$. Since a section of a sheaf is determined by its germs in each stalk, $a(s)neq b(s)$ means there exists some $xin U$ such that $a(s)_xneq b(s)_x$ in $G_x$. But $a(s)_x=a_x(s_x)$ and $b(s)_s=b_x(s_x)$, so this proves that $a_xneq b_x$. That is, there is some stalk on which $a$ and $b$ are not equal.
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
add a comment |Â
up vote
6
down vote
accepted
Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:Fto G$ are morphisms which are not equal. Then there is some open $Usubseteq X$ and some $sin F(U)$ such that $a(s)neq b(s)$. Since a section of a sheaf is determined by its germs in each stalk, $a(s)neq b(s)$ means there exists some $xin U$ such that $a(s)_xneq b(s)_x$ in $G_x$. But $a(s)_x=a_x(s_x)$ and $b(s)_s=b_x(s_x)$, so this proves that $a_xneq b_x$. That is, there is some stalk on which $a$ and $b$ are not equal.
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:Fto G$ are morphisms which are not equal. Then there is some open $Usubseteq X$ and some $sin F(U)$ such that $a(s)neq b(s)$. Since a section of a sheaf is determined by its germs in each stalk, $a(s)neq b(s)$ means there exists some $xin U$ such that $a(s)_xneq b(s)_x$ in $G_x$. But $a(s)_x=a_x(s_x)$ and $b(s)_s=b_x(s_x)$, so this proves that $a_xneq b_x$. That is, there is some stalk on which $a$ and $b$ are not equal.
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
Yes: to prove two morphisms of sheaves are equal, it suffices to prove they induce the same maps on each stalk. Explicitly, let us suppose $F$ and $G$ are sheaves on a space $X$ and $a,b:Fto G$ are morphisms which are not equal. Then there is some open $Usubseteq X$ and some $sin F(U)$ such that $a(s)neq b(s)$. Since a section of a sheaf is determined by its germs in each stalk, $a(s)neq b(s)$ means there exists some $xin U$ such that $a(s)_xneq b(s)_x$ in $G_x$. But $a(s)_x=a_x(s_x)$ and $b(s)_s=b_x(s_x)$, so this proves that $a_xneq b_x$. That is, there is some stalk on which $a$ and $b$ are not equal.
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
answered Jul 31 at 21:27
Eric Wofsey
161k12188297
161k12188297
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%2f2868487%2fhartshorne-exercise-ii-1-18%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
If it's the sheafification part of the definition of $f^-1$ that's tripping you up, you might try first proving $f_* : PSh_Y to PSh_X$ and $f^circ : PSh_X to PSh_Y$ are adjoint; then use that sheafification is also a left adjoint of the inclusion $Sh_X to PSh_X$.
– Daniel Schepler
Jul 31 at 21:35
@DanielSchepler Ah, yes, that came to mind but I didn't went further with it. Thanks.
– Pedro Tamaroff♦
Jul 31 at 21:42