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Confusion about elements of the stalk of the direct image sheaf
Clash Royale CLAN TAG#URR8PPP
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$DeclareMathOperatorFmathcalF$Let $F$ be a sheaf on $X$ and $pi : X rightarrow Y$ a continuous map. Then the the direct image sheaf $pi_* F$ is a sheaf on $Y$. An explicit definition of the stalk the sheaf $F$ at point $p in X$ is as follows:
$$ F_p = (f, U) mid p in U, , , f in F(U) / sim $$
where $(f, U) sim (g, V)$ if and only if there exists an open $W subset U cap V$ such that $f|_W = g|_W$. Now, the stalk of $pi_* F$ at point $q = pi(p) in Y$ is as follows:
$$ beginalign* (pi_* F)_q &= (h, Z) mid q in Z, , , h in pi_* F(Z) = F(pi^-1(Z)) / sim \ &stackrelcolorred?= (h, pi^-1(Z)) mid p in pi^-1(Z), , , h in F(pi^-1(Z)) / sim endalign* $$
but then since $h in F(pi^-1(Z))$, it's e.g. a continuous function in some subset of $X$ (say the sheaves are of continuous functions). Hasn't something gone wrong here? Surely if I'm talking about the stalk of $pi_* F$, then shouldn't the function be in $Y$, not $X$? I know that there's a natural map between $(pi_* F)_q$ and $F_p$ so this likely has something do with my question, but all I've done above is write out what the stalk $(pi_* F)_q$ is explicitly, and there already seems to be some contradiction regarding the function $h$.
Thank you for any help.
Edit: Having looked at the above, I don't think it's valid me jumping to the second line because this is no longer correct as an explicit realisation of the stalk of $pi_* F$, since inside the curly brackets I'm taking sections in $F$ whereas I should be taking them in $pi_* F$.
However I'm still confused; doesn't $f in pi_*F(U)$ mean that $f$ is, say, a continuous function in $Y$ (because $pi_*F$ is a sheaf over $Y$), but then simultaneously $f in F(pi^-1(U))$ is a continuous function in $X$ because $F$ is a sheaf over $X$? I feel I'm misunderstanding something really obvious here.
Edit 2: Maybe this isn't relevant. $F$ being a sheaf on $X$ means its arguments should be open sets in $X$, but it doesn't actually say anything about what the sections have to be?
general-topology sheaf-theory
edited Jul 27 at 10:54
Armando j18eos
2,35711125
asked Jul 25 at 17:45
mathphys
961415
 |Â
show 2 more comments
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0
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$DeclareMathOperatorFmathcalF$Let $F$ be a sheaf on $X$ and $pi : X rightarrow Y$ a continuous map. Then the the direct image sheaf $pi_* F$ is a sheaf on $Y$. An explicit definition of the stalk the sheaf $F$ at point $p in X$ is as follows:
$$ F_p = (f, U) mid p in U, , , f in F(U) / sim $$
where $(f, U) sim (g, V)$ if and only if there exists an open $W subset U cap V$ such that $f|_W = g|_W$. Now, the stalk of $pi_* F$ at point $q = pi(p) in Y$ is as follows:
$$ beginalign* (pi_* F)_q &= (h, Z) mid q in Z, , , h in pi_* F(Z) = F(pi^-1(Z)) / sim \ &stackrelcolorred?= (h, pi^-1(Z)) mid p in pi^-1(Z), , , h in F(pi^-1(Z)) / sim endalign* $$
but then since $h in F(pi^-1(Z))$, it's e.g. a continuous function in some subset of $X$ (say the sheaves are of continuous functions). Hasn't something gone wrong here? Surely if I'm talking about the stalk of $pi_* F$, then shouldn't the function be in $Y$, not $X$? I know that there's a natural map between $(pi_* F)_q$ and $F_p$ so this likely has something do with my question, but all I've done above is write out what the stalk $(pi_* F)_q$ is explicitly, and there already seems to be some contradiction regarding the function $h$.
Thank you for any help.
Edit: Having looked at the above, I don't think it's valid me jumping to the second line because this is no longer correct as an explicit realisation of the stalk of $pi_* F$, since inside the curly brackets I'm taking sections in $F$ whereas I should be taking them in $pi_* F$.
However I'm still confused; doesn't $f in pi_*F(U)$ mean that $f$ is, say, a continuous function in $Y$ (because $pi_*F$ is a sheaf over $Y$), but then simultaneously $f in F(pi^-1(U))$ is a continuous function in $X$ because $F$ is a sheaf over $X$? I feel I'm misunderstanding something really obvious here.
Edit 2: Maybe this isn't relevant. $F$ being a sheaf on $X$ means its arguments should be open sets in $X$, but it doesn't actually say anything about what the sections have to be?
general-topology sheaf-theory
edited Jul 27 at 10:54
Armando j18eos
2,35711125
asked Jul 25 at 17:45
mathphys
961415
3
Don't forget that for $pin Y$ there may be several $pin X$ with $pi(x)=y$, (or perhaps none).
– Lord Shark the Unknown
Jul 25 at 17:49
Do you by any chance mean for $q in Y$, there may be several $p in X$ with $pi(p)=q$? Also, how could there be none if we've said initially that $pi(p)=q$?
– mathphys
Jul 25 at 17:53
1
The direct image sheaf will still have a stalk at $q$ even if $pi^-1(q)=emptyset$,
– Lord Shark the Unknown
Jul 25 at 17:54
1
The stalks of $pi_*mathcal F$ are equivalence classes of certain sections of $mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them.
– Lord Shark the Unknown
Jul 25 at 18:11
1
Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $mathcalF$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.)
– Alex Wertheim
Jul 25 at 21:07
 |Â
show 2 more comments
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0
down vote
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0
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$DeclareMathOperatorFmathcalF$Let $F$ be a sheaf on $X$ and $pi : X rightarrow Y$ a continuous map. Then the the direct image sheaf $pi_* F$ is a sheaf on $Y$. An explicit definition of the stalk the sheaf $F$ at point $p in X$ is as follows:
$$ F_p = (f, U) mid p in U, , , f in F(U) / sim $$
where $(f, U) sim (g, V)$ if and only if there exists an open $W subset U cap V$ such that $f|_W = g|_W$. Now, the stalk of $pi_* F$ at point $q = pi(p) in Y$ is as follows:
$$ beginalign* (pi_* F)_q &= (h, Z) mid q in Z, , , h in pi_* F(Z) = F(pi^-1(Z)) / sim \ &stackrelcolorred?= (h, pi^-1(Z)) mid p in pi^-1(Z), , , h in F(pi^-1(Z)) / sim endalign* $$
but then since $h in F(pi^-1(Z))$, it's e.g. a continuous function in some subset of $X$ (say the sheaves are of continuous functions). Hasn't something gone wrong here? Surely if I'm talking about the stalk of $pi_* F$, then shouldn't the function be in $Y$, not $X$? I know that there's a natural map between $(pi_* F)_q$ and $F_p$ so this likely has something do with my question, but all I've done above is write out what the stalk $(pi_* F)_q$ is explicitly, and there already seems to be some contradiction regarding the function $h$.
Thank you for any help.
Edit: Having looked at the above, I don't think it's valid me jumping to the second line because this is no longer correct as an explicit realisation of the stalk of $pi_* F$, since inside the curly brackets I'm taking sections in $F$ whereas I should be taking them in $pi_* F$.
However I'm still confused; doesn't $f in pi_*F(U)$ mean that $f$ is, say, a continuous function in $Y$ (because $pi_*F$ is a sheaf over $Y$), but then simultaneously $f in F(pi^-1(U))$ is a continuous function in $X$ because $F$ is a sheaf over $X$? I feel I'm misunderstanding something really obvious here.
Edit 2: Maybe this isn't relevant. $F$ being a sheaf on $X$ means its arguments should be open sets in $X$, but it doesn't actually say anything about what the sections have to be?
general-topology sheaf-theory
edited Jul 27 at 10:54
Armando j18eos
2,35711125
asked Jul 25 at 17:45
mathphys
961415
$DeclareMathOperatorFmathcalF$Let $F$ be a sheaf on $X$ and $pi : X rightarrow Y$ a continuous map. Then the the direct image sheaf $pi_* F$ is a sheaf on $Y$. An explicit definition of the stalk the sheaf $F$ at point $p in X$ is as follows:
$$ F_p = (f, U) mid p in U, , , f in F(U) / sim $$
where $(f, U) sim (g, V)$ if and only if there exists an open $W subset U cap V$ such that $f|_W = g|_W$. Now, the stalk of $pi_* F$ at point $q = pi(p) in Y$ is as follows:
$$ beginalign* (pi_* F)_q &= (h, Z) mid q in Z, , , h in pi_* F(Z) = F(pi^-1(Z)) / sim \ &stackrelcolorred?= (h, pi^-1(Z)) mid p in pi^-1(Z), , , h in F(pi^-1(Z)) / sim endalign* $$
but then since $h in F(pi^-1(Z))$, it's e.g. a continuous function in some subset of $X$ (say the sheaves are of continuous functions). Hasn't something gone wrong here? Surely if I'm talking about the stalk of $pi_* F$, then shouldn't the function be in $Y$, not $X$? I know that there's a natural map between $(pi_* F)_q$ and $F_p$ so this likely has something do with my question, but all I've done above is write out what the stalk $(pi_* F)_q$ is explicitly, and there already seems to be some contradiction regarding the function $h$.
Thank you for any help.
Edit: Having looked at the above, I don't think it's valid me jumping to the second line because this is no longer correct as an explicit realisation of the stalk of $pi_* F$, since inside the curly brackets I'm taking sections in $F$ whereas I should be taking them in $pi_* F$.
However I'm still confused; doesn't $f in pi_*F(U)$ mean that $f$ is, say, a continuous function in $Y$ (because $pi_*F$ is a sheaf over $Y$), but then simultaneously $f in F(pi^-1(U))$ is a continuous function in $X$ because $F$ is a sheaf over $X$? I feel I'm misunderstanding something really obvious here.
Edit 2: Maybe this isn't relevant. $F$ being a sheaf on $X$ means its arguments should be open sets in $X$, but it doesn't actually say anything about what the sections have to be?
general-topology sheaf-theory
edited Jul 27 at 10:54
Armando j18eos
2,35711125
asked Jul 25 at 17:45
mathphys
961415
edited Jul 27 at 10:54
Armando j18eos
2,35711125
edited Jul 27 at 10:54
Armando j18eos
2,35711125
edited Jul 27 at 10:54
Armando j18eos
2,35711125
2,35711125
asked Jul 25 at 17:45
mathphys
961415
asked Jul 25 at 17:45
mathphys
961415
asked Jul 25 at 17:45
mathphys
961415
961415
3
Don't forget that for $pin Y$ there may be several $pin X$ with $pi(x)=y$, (or perhaps none).
– Lord Shark the Unknown
Jul 25 at 17:49
Do you by any chance mean for $q in Y$, there may be several $p in X$ with $pi(p)=q$? Also, how could there be none if we've said initially that $pi(p)=q$?
– mathphys
Jul 25 at 17:53
1
The direct image sheaf will still have a stalk at $q$ even if $pi^-1(q)=emptyset$,
– Lord Shark the Unknown
Jul 25 at 17:54
1
The stalks of $pi_*mathcal F$ are equivalence classes of certain sections of $mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them.
– Lord Shark the Unknown
Jul 25 at 18:11
1
Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $mathcalF$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.)
– Alex Wertheim
Jul 25 at 21:07
 |Â
show 2 more comments
3
Don't forget that for $pin Y$ there may be several $pin X$ with $pi(x)=y$, (or perhaps none).
– Lord Shark the Unknown
Jul 25 at 17:49
Do you by any chance mean for $q in Y$, there may be several $p in X$ with $pi(p)=q$? Also, how could there be none if we've said initially that $pi(p)=q$?
– mathphys
Jul 25 at 17:53
1
The direct image sheaf will still have a stalk at $q$ even if $pi^-1(q)=emptyset$,
– Lord Shark the Unknown
Jul 25 at 17:54
1
The stalks of $pi_*mathcal F$ are equivalence classes of certain sections of $mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them.
– Lord Shark the Unknown
Jul 25 at 18:11
1
Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $mathcalF$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.)
– Alex Wertheim
Jul 25 at 21:07
3
3
Don't forget that for $pin Y$ there may be several $pin X$ with $pi(x)=y$, (or perhaps none).
– Lord Shark the Unknown
Jul 25 at 17:49
Don't forget that for $pin Y$ there may be several $pin X$ with $pi(x)=y$, (or perhaps none).
– Lord Shark the Unknown
Jul 25 at 17:49
Do you by any chance mean for $q in Y$, there may be several $p in X$ with $pi(p)=q$? Also, how could there be none if we've said initially that $pi(p)=q$?
– mathphys
Jul 25 at 17:53
Do you by any chance mean for $q in Y$, there may be several $p in X$ with $pi(p)=q$? Also, how could there be none if we've said initially that $pi(p)=q$?
– mathphys
Jul 25 at 17:53
1
1
The direct image sheaf will still have a stalk at $q$ even if $pi^-1(q)=emptyset$,
– Lord Shark the Unknown
Jul 25 at 17:54
The direct image sheaf will still have a stalk at $q$ even if $pi^-1(q)=emptyset$,
– Lord Shark the Unknown
Jul 25 at 17:54
1
1
The stalks of $pi_*mathcal F$ are equivalence classes of certain sections of $mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them.
– Lord Shark the Unknown
Jul 25 at 18:11
The stalks of $pi_*mathcal F$ are equivalence classes of certain sections of $mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them.
– Lord Shark the Unknown
Jul 25 at 18:11
1
1
Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $mathcalF$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.)
– Alex Wertheim
Jul 25 at 21:07
Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $mathcalF$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.)
– Alex Wertheim
Jul 25 at 21:07
 |Â
show 2 more comments
1 Answer
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Let $f:Xto Y$ be a continuous map of topological spaces, let $mathcalS$ be the sheaf of continuous functions on $X$ with values in $mathbbK$ with the natural topology, where $mathbbK$ is either $mathbbR$ the field of real numbers or $mathbbC$ the field of complex numbers.
For exact:
beginequation
forall Usubseteq X,textopen,,mathcalS(U)=c:UtomathbbKmid c,textis continuous;
endequation
because $mathcalS$ is a sheaf of rings on $X$, the pair $(X,mathcalS)$ is called ringed space.
By definition:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)stackreldef.=mathcalS(f^-1(V)),
endequation
that is: $pi_*mathcalS$ is a sheaf of rings on $Y$; the elements of $f_*mathcalS(V)$ can be identified with functions on $V$, with values in $mathbbK$ which admit a factorization via $f$. In other words, $f_*mathcalS$ is a subsheaf of the sheaf of functions on $Y$ with values in $mathbbK$.
Remark. For clarity:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)=d:VtomathbbKmidexists cinmathcalS(f^-1(V)),texts.t.,d=ccirc f,
endequation
and one can not state that $d$ is continuous on $V$. $Diamond$
Let
beginequation
widetildeX=X_displaystyle/(xsim yiff f(x)=f(y)),
endequation
that is $widetildeX$ is the quotient set of $X$ where the points with the same image via $f$ are identied; let $pi:XtowidetildeX$ be the canonical projection, and let $varphi:widetildeXto Y$ be the unique function such that
beginequation
f=varphicircpi.
endequation
One knows that $varphi$ is an injective map; considering on $widetildeX$ the quotient topology, one has that $pi$ and $varphi$ are continuous maps and
beginequation
f_*mathcalS=(varphicircpi)_*mathcalS=varphi_*left(pi_*mathcalSright).
endequation
Because $f(X)=varphileft(widetildeXright)=Z$ is a subset of $Y$ and $varphi$ is injective, one knows that:
beginequation
forall yin Y,,left(f_*mathcalSright)_y=begincases
0iff ynotinoverlineZ\
left(pi_*mathcalSright)_ziffvarphi(z)=yin Z\
?iff yinoverlineZsetminus Z
endcases.
endequation
Example. Let
beginequation
X=a,b,c,,mathcalT=emptyset,b,c,b,c,X;
endequation
$a$ is a closed point and $b$ is an open point of $X$. Let
beginequation
mathcalF(emptyset)=0,,mathcalF(b)=mathbbZ,
endequation
this is a sheaf on $b$ with the topology of subspace of $(X,mathcalT)$; easily one has:
beginequation
i_*mathcalF(X)=mathbbZRightarrowleft(i_*mathcalFright)_a=mathbbZ,\
left(i_*mathcalFright)_b=mathbbZ,\
i_*mathcalF(c)=0,i_*mathcalF(b,c)=mathbbZRightarrowleft(i_*mathcalFright)_c=0
endequation
where $i:bhookrightarrow X$ is the inclusion and $ainoverlinebsetminusb$. Instead, let $mathcalG$ be the analogous sheaf on $a$; one has that $j_*mathcalG$ is the skyscraper sheaf on $X$ with stalk $mathbbZ$ at $a$, where $j:ato X$ is the inclusion. $triangle$
So, also if one computes what is $left(pi_*mathcalSright)_z$ for any $zinwidetildeX$, in general one can state nothing about $left(f_*mathcalSright)_y$ for $yinoverlineZsetminus Z$ a priori.
answered Jul 27 at 10:52
Armando j18eos
2,35711125
add a comment |Â
1 Answer
1
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1 Answer
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active
oldest
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active
oldest
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active
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up vote
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Let $f:Xto Y$ be a continuous map of topological spaces, let $mathcalS$ be the sheaf of continuous functions on $X$ with values in $mathbbK$ with the natural topology, where $mathbbK$ is either $mathbbR$ the field of real numbers or $mathbbC$ the field of complex numbers.
For exact:
beginequation
forall Usubseteq X,textopen,,mathcalS(U)=c:UtomathbbKmid c,textis continuous;
endequation
because $mathcalS$ is a sheaf of rings on $X$, the pair $(X,mathcalS)$ is called ringed space.
By definition:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)stackreldef.=mathcalS(f^-1(V)),
endequation
that is: $pi_*mathcalS$ is a sheaf of rings on $Y$; the elements of $f_*mathcalS(V)$ can be identified with functions on $V$, with values in $mathbbK$ which admit a factorization via $f$. In other words, $f_*mathcalS$ is a subsheaf of the sheaf of functions on $Y$ with values in $mathbbK$.
Remark. For clarity:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)=d:VtomathbbKmidexists cinmathcalS(f^-1(V)),texts.t.,d=ccirc f,
endequation
and one can not state that $d$ is continuous on $V$. $Diamond$
Let
beginequation
widetildeX=X_displaystyle/(xsim yiff f(x)=f(y)),
endequation
that is $widetildeX$ is the quotient set of $X$ where the points with the same image via $f$ are identied; let $pi:XtowidetildeX$ be the canonical projection, and let $varphi:widetildeXto Y$ be the unique function such that
beginequation
f=varphicircpi.
endequation
One knows that $varphi$ is an injective map; considering on $widetildeX$ the quotient topology, one has that $pi$ and $varphi$ are continuous maps and
beginequation
f_*mathcalS=(varphicircpi)_*mathcalS=varphi_*left(pi_*mathcalSright).
endequation
Because $f(X)=varphileft(widetildeXright)=Z$ is a subset of $Y$ and $varphi$ is injective, one knows that:
beginequation
forall yin Y,,left(f_*mathcalSright)_y=begincases
0iff ynotinoverlineZ\
left(pi_*mathcalSright)_ziffvarphi(z)=yin Z\
?iff yinoverlineZsetminus Z
endcases.
endequation
Example. Let
beginequation
X=a,b,c,,mathcalT=emptyset,b,c,b,c,X;
endequation
$a$ is a closed point and $b$ is an open point of $X$. Let
beginequation
mathcalF(emptyset)=0,,mathcalF(b)=mathbbZ,
endequation
this is a sheaf on $b$ with the topology of subspace of $(X,mathcalT)$; easily one has:
beginequation
i_*mathcalF(X)=mathbbZRightarrowleft(i_*mathcalFright)_a=mathbbZ,\
left(i_*mathcalFright)_b=mathbbZ,\
i_*mathcalF(c)=0,i_*mathcalF(b,c)=mathbbZRightarrowleft(i_*mathcalFright)_c=0
endequation
where $i:bhookrightarrow X$ is the inclusion and $ainoverlinebsetminusb$. Instead, let $mathcalG$ be the analogous sheaf on $a$; one has that $j_*mathcalG$ is the skyscraper sheaf on $X$ with stalk $mathbbZ$ at $a$, where $j:ato X$ is the inclusion. $triangle$
So, also if one computes what is $left(pi_*mathcalSright)_z$ for any $zinwidetildeX$, in general one can state nothing about $left(f_*mathcalSright)_y$ for $yinoverlineZsetminus Z$ a priori.
answered Jul 27 at 10:52
Armando j18eos
2,35711125
add a comment |Â
up vote
0
down vote
Let $f:Xto Y$ be a continuous map of topological spaces, let $mathcalS$ be the sheaf of continuous functions on $X$ with values in $mathbbK$ with the natural topology, where $mathbbK$ is either $mathbbR$ the field of real numbers or $mathbbC$ the field of complex numbers.
For exact:
beginequation
forall Usubseteq X,textopen,,mathcalS(U)=c:UtomathbbKmid c,textis continuous;
endequation
because $mathcalS$ is a sheaf of rings on $X$, the pair $(X,mathcalS)$ is called ringed space.
By definition:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)stackreldef.=mathcalS(f^-1(V)),
endequation
that is: $pi_*mathcalS$ is a sheaf of rings on $Y$; the elements of $f_*mathcalS(V)$ can be identified with functions on $V$, with values in $mathbbK$ which admit a factorization via $f$. In other words, $f_*mathcalS$ is a subsheaf of the sheaf of functions on $Y$ with values in $mathbbK$.
Remark. For clarity:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)=d:VtomathbbKmidexists cinmathcalS(f^-1(V)),texts.t.,d=ccirc f,
endequation
and one can not state that $d$ is continuous on $V$. $Diamond$
Let
beginequation
widetildeX=X_displaystyle/(xsim yiff f(x)=f(y)),
endequation
that is $widetildeX$ is the quotient set of $X$ where the points with the same image via $f$ are identied; let $pi:XtowidetildeX$ be the canonical projection, and let $varphi:widetildeXto Y$ be the unique function such that
beginequation
f=varphicircpi.
endequation
One knows that $varphi$ is an injective map; considering on $widetildeX$ the quotient topology, one has that $pi$ and $varphi$ are continuous maps and
beginequation
f_*mathcalS=(varphicircpi)_*mathcalS=varphi_*left(pi_*mathcalSright).
endequation
Because $f(X)=varphileft(widetildeXright)=Z$ is a subset of $Y$ and $varphi$ is injective, one knows that:
beginequation
forall yin Y,,left(f_*mathcalSright)_y=begincases
0iff ynotinoverlineZ\
left(pi_*mathcalSright)_ziffvarphi(z)=yin Z\
?iff yinoverlineZsetminus Z
endcases.
endequation
Example. Let
beginequation
X=a,b,c,,mathcalT=emptyset,b,c,b,c,X;
endequation
$a$ is a closed point and $b$ is an open point of $X$. Let
beginequation
mathcalF(emptyset)=0,,mathcalF(b)=mathbbZ,
endequation
this is a sheaf on $b$ with the topology of subspace of $(X,mathcalT)$; easily one has:
beginequation
i_*mathcalF(X)=mathbbZRightarrowleft(i_*mathcalFright)_a=mathbbZ,\
left(i_*mathcalFright)_b=mathbbZ,\
i_*mathcalF(c)=0,i_*mathcalF(b,c)=mathbbZRightarrowleft(i_*mathcalFright)_c=0
endequation
where $i:bhookrightarrow X$ is the inclusion and $ainoverlinebsetminusb$. Instead, let $mathcalG$ be the analogous sheaf on $a$; one has that $j_*mathcalG$ is the skyscraper sheaf on $X$ with stalk $mathbbZ$ at $a$, where $j:ato X$ is the inclusion. $triangle$
So, also if one computes what is $left(pi_*mathcalSright)_z$ for any $zinwidetildeX$, in general one can state nothing about $left(f_*mathcalSright)_y$ for $yinoverlineZsetminus Z$ a priori.
answered Jul 27 at 10:52
Armando j18eos
2,35711125
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $f:Xto Y$ be a continuous map of topological spaces, let $mathcalS$ be the sheaf of continuous functions on $X$ with values in $mathbbK$ with the natural topology, where $mathbbK$ is either $mathbbR$ the field of real numbers or $mathbbC$ the field of complex numbers.
For exact:
beginequation
forall Usubseteq X,textopen,,mathcalS(U)=c:UtomathbbKmid c,textis continuous;
endequation
because $mathcalS$ is a sheaf of rings on $X$, the pair $(X,mathcalS)$ is called ringed space.
By definition:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)stackreldef.=mathcalS(f^-1(V)),
endequation
that is: $pi_*mathcalS$ is a sheaf of rings on $Y$; the elements of $f_*mathcalS(V)$ can be identified with functions on $V$, with values in $mathbbK$ which admit a factorization via $f$. In other words, $f_*mathcalS$ is a subsheaf of the sheaf of functions on $Y$ with values in $mathbbK$.
Remark. For clarity:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)=d:VtomathbbKmidexists cinmathcalS(f^-1(V)),texts.t.,d=ccirc f,
endequation
and one can not state that $d$ is continuous on $V$. $Diamond$
Let
beginequation
widetildeX=X_displaystyle/(xsim yiff f(x)=f(y)),
endequation
that is $widetildeX$ is the quotient set of $X$ where the points with the same image via $f$ are identied; let $pi:XtowidetildeX$ be the canonical projection, and let $varphi:widetildeXto Y$ be the unique function such that
beginequation
f=varphicircpi.
endequation
One knows that $varphi$ is an injective map; considering on $widetildeX$ the quotient topology, one has that $pi$ and $varphi$ are continuous maps and
beginequation
f_*mathcalS=(varphicircpi)_*mathcalS=varphi_*left(pi_*mathcalSright).
endequation
Because $f(X)=varphileft(widetildeXright)=Z$ is a subset of $Y$ and $varphi$ is injective, one knows that:
beginequation
forall yin Y,,left(f_*mathcalSright)_y=begincases
0iff ynotinoverlineZ\
left(pi_*mathcalSright)_ziffvarphi(z)=yin Z\
?iff yinoverlineZsetminus Z
endcases.
endequation
Example. Let
beginequation
X=a,b,c,,mathcalT=emptyset,b,c,b,c,X;
endequation
$a$ is a closed point and $b$ is an open point of $X$. Let
beginequation
mathcalF(emptyset)=0,,mathcalF(b)=mathbbZ,
endequation
this is a sheaf on $b$ with the topology of subspace of $(X,mathcalT)$; easily one has:
beginequation
i_*mathcalF(X)=mathbbZRightarrowleft(i_*mathcalFright)_a=mathbbZ,\
left(i_*mathcalFright)_b=mathbbZ,\
i_*mathcalF(c)=0,i_*mathcalF(b,c)=mathbbZRightarrowleft(i_*mathcalFright)_c=0
endequation
where $i:bhookrightarrow X$ is the inclusion and $ainoverlinebsetminusb$. Instead, let $mathcalG$ be the analogous sheaf on $a$; one has that $j_*mathcalG$ is the skyscraper sheaf on $X$ with stalk $mathbbZ$ at $a$, where $j:ato X$ is the inclusion. $triangle$
So, also if one computes what is $left(pi_*mathcalSright)_z$ for any $zinwidetildeX$, in general one can state nothing about $left(f_*mathcalSright)_y$ for $yinoverlineZsetminus Z$ a priori.
answered Jul 27 at 10:52
Armando j18eos
2,35711125
Let $f:Xto Y$ be a continuous map of topological spaces, let $mathcalS$ be the sheaf of continuous functions on $X$ with values in $mathbbK$ with the natural topology, where $mathbbK$ is either $mathbbR$ the field of real numbers or $mathbbC$ the field of complex numbers.
For exact:
beginequation
forall Usubseteq X,textopen,,mathcalS(U)=c:UtomathbbKmid c,textis continuous;
endequation
because $mathcalS$ is a sheaf of rings on $X$, the pair $(X,mathcalS)$ is called ringed space.
By definition:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)stackreldef.=mathcalS(f^-1(V)),
endequation
that is: $pi_*mathcalS$ is a sheaf of rings on $Y$; the elements of $f_*mathcalS(V)$ can be identified with functions on $V$, with values in $mathbbK$ which admit a factorization via $f$. In other words, $f_*mathcalS$ is a subsheaf of the sheaf of functions on $Y$ with values in $mathbbK$.
Remark. For clarity:
beginequation
forall Vsubseteq Y,textopen,,f_*mathcalS(V)=d:VtomathbbKmidexists cinmathcalS(f^-1(V)),texts.t.,d=ccirc f,
endequation
and one can not state that $d$ is continuous on $V$. $Diamond$
Let
beginequation
widetildeX=X_displaystyle/(xsim yiff f(x)=f(y)),
endequation
that is $widetildeX$ is the quotient set of $X$ where the points with the same image via $f$ are identied; let $pi:XtowidetildeX$ be the canonical projection, and let $varphi:widetildeXto Y$ be the unique function such that
beginequation
f=varphicircpi.
endequation
One knows that $varphi$ is an injective map; considering on $widetildeX$ the quotient topology, one has that $pi$ and $varphi$ are continuous maps and
beginequation
f_*mathcalS=(varphicircpi)_*mathcalS=varphi_*left(pi_*mathcalSright).
endequation
Because $f(X)=varphileft(widetildeXright)=Z$ is a subset of $Y$ and $varphi$ is injective, one knows that:
beginequation
forall yin Y,,left(f_*mathcalSright)_y=begincases
0iff ynotinoverlineZ\
left(pi_*mathcalSright)_ziffvarphi(z)=yin Z\
?iff yinoverlineZsetminus Z
endcases.
endequation
Example. Let
beginequation
X=a,b,c,,mathcalT=emptyset,b,c,b,c,X;
endequation
$a$ is a closed point and $b$ is an open point of $X$. Let
beginequation
mathcalF(emptyset)=0,,mathcalF(b)=mathbbZ,
endequation
this is a sheaf on $b$ with the topology of subspace of $(X,mathcalT)$; easily one has:
beginequation
i_*mathcalF(X)=mathbbZRightarrowleft(i_*mathcalFright)_a=mathbbZ,\
left(i_*mathcalFright)_b=mathbbZ,\
i_*mathcalF(c)=0,i_*mathcalF(b,c)=mathbbZRightarrowleft(i_*mathcalFright)_c=0
endequation
where $i:bhookrightarrow X$ is the inclusion and $ainoverlinebsetminusb$. Instead, let $mathcalG$ be the analogous sheaf on $a$; one has that $j_*mathcalG$ is the skyscraper sheaf on $X$ with stalk $mathbbZ$ at $a$, where $j:ato X$ is the inclusion. $triangle$
So, also if one computes what is $left(pi_*mathcalSright)_z$ for any $zinwidetildeX$, in general one can state nothing about $left(f_*mathcalSright)_y$ for $yinoverlineZsetminus Z$ a priori.
answered Jul 27 at 10:52
Armando j18eos
2,35711125
answered Jul 27 at 10:52
Armando j18eos
2,35711125
answered Jul 27 at 10:52
Armando j18eos
2,35711125
answered Jul 27 at 10:52
Armando j18eos
2,35711125
2,35711125
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3
Don't forget that for $pin Y$ there may be several $pin X$ with $pi(x)=y$, (or perhaps none).
– Lord Shark the Unknown
Jul 25 at 17:49
Do you by any chance mean for $q in Y$, there may be several $p in X$ with $pi(p)=q$? Also, how could there be none if we've said initially that $pi(p)=q$?
– mathphys
Jul 25 at 17:53
1
The direct image sheaf will still have a stalk at $q$ even if $pi^-1(q)=emptyset$,
– Lord Shark the Unknown
Jul 25 at 17:54
1
The stalks of $pi_*mathcal F$ are equivalence classes of certain sections of $mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them.
– Lord Shark the Unknown
Jul 25 at 18:11
1
Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $mathcalF$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.)
– Alex Wertheim
Jul 25 at 21:07