Stalks of the pushforward sheaf.
$begingroup$
I know that in general, the stalks of the pushforward of a sheaf need not be the same as the original stalk. That is, $(f_{*}mathcal{F})_{f(p)}=mathcal{F}_p$ is not true in general. But when $X subset Y$ and $f$ is the inclusion map, is it true that the above holds for $p in X$ and that the stalks are zero(or the terminal object) for points not in X?
I am asking this because I was able to prove this for any subset, but in Hartshorne Chapter 2, exercise 19, he asks you to prove that this holds for closed subsets and that a different construction(denoted $f_{!}mathcal{F}$) gives these stalks for open subsets.
algebraic-geometry sheaf-theory
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
$endgroup$
add a comment |
$begingroup$
I know that in general, the stalks of the pushforward of a sheaf need not be the same as the original stalk. That is, $(f_{*}mathcal{F})_{f(p)}=mathcal{F}_p$ is not true in general. But when $X subset Y$ and $f$ is the inclusion map, is it true that the above holds for $p in X$ and that the stalks are zero(or the terminal object) for points not in X?
I am asking this because I was able to prove this for any subset, but in Hartshorne Chapter 2, exercise 19, he asks you to prove that this holds for closed subsets and that a different construction(denoted $f_{!}mathcal{F}$) gives these stalks for open subsets.
algebraic-geometry sheaf-theory
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
$endgroup$
add a comment |
$begingroup$
I know that in general, the stalks of the pushforward of a sheaf need not be the same as the original stalk. That is, $(f_{*}mathcal{F})_{f(p)}=mathcal{F}_p$ is not true in general. But when $X subset Y$ and $f$ is the inclusion map, is it true that the above holds for $p in X$ and that the stalks are zero(or the terminal object) for points not in X?
I am asking this because I was able to prove this for any subset, but in Hartshorne Chapter 2, exercise 19, he asks you to prove that this holds for closed subsets and that a different construction(denoted $f_{!}mathcal{F}$) gives these stalks for open subsets.
algebraic-geometry sheaf-theory
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
$endgroup$
I know that in general, the stalks of the pushforward of a sheaf need not be the same as the original stalk. That is, $(f_{*}mathcal{F})_{f(p)}=mathcal{F}_p$ is not true in general. But when $X subset Y$ and $f$ is the inclusion map, is it true that the above holds for $p in X$ and that the stalks are zero(or the terminal object) for points not in X?
I am asking this because I was able to prove this for any subset, but in Hartshorne Chapter 2, exercise 19, he asks you to prove that this holds for closed subsets and that a different construction(denoted $f_{!}mathcal{F}$) gives these stalks for open subsets.
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
asked Dec 22 '18 at 9:19
Jehu314Jehu314
1548
1548
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No it is not true, take $Y$ to be the unit disc in $mathbb{C}$ and $Xsubset Y$ to be the punctured unit disc and $mathcal{F}$ to be the constant sheaf of fiber say $mathbb{Z}$ over $X$. The inclusion will be denoted by $i$, and $0$ will be the origin of the disc.
Then the fiber of $i_*mathcal{F}$ at $0$ will be... $mathbb{Z}$.
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
$endgroup$
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
1
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049243%2fstalks-of-the-pushforward-sheaf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No it is not true, take $Y$ to be the unit disc in $mathbb{C}$ and $Xsubset Y$ to be the punctured unit disc and $mathcal{F}$ to be the constant sheaf of fiber say $mathbb{Z}$ over $X$. The inclusion will be denoted by $i$, and $0$ will be the origin of the disc.
Then the fiber of $i_*mathcal{F}$ at $0$ will be... $mathbb{Z}$.
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
$endgroup$
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
1
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
add a comment |
$begingroup$
No it is not true, take $Y$ to be the unit disc in $mathbb{C}$ and $Xsubset Y$ to be the punctured unit disc and $mathcal{F}$ to be the constant sheaf of fiber say $mathbb{Z}$ over $X$. The inclusion will be denoted by $i$, and $0$ will be the origin of the disc.
Then the fiber of $i_*mathcal{F}$ at $0$ will be... $mathbb{Z}$.
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
$endgroup$
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
1
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
add a comment |
$begingroup$
No it is not true, take $Y$ to be the unit disc in $mathbb{C}$ and $Xsubset Y$ to be the punctured unit disc and $mathcal{F}$ to be the constant sheaf of fiber say $mathbb{Z}$ over $X$. The inclusion will be denoted by $i$, and $0$ will be the origin of the disc.
Then the fiber of $i_*mathcal{F}$ at $0$ will be... $mathbb{Z}$.
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
$endgroup$
No it is not true, take $Y$ to be the unit disc in $mathbb{C}$ and $Xsubset Y$ to be the punctured unit disc and $mathcal{F}$ to be the constant sheaf of fiber say $mathbb{Z}$ over $X$. The inclusion will be denoted by $i$, and $0$ will be the origin of the disc.
Then the fiber of $i_*mathcal{F}$ at $0$ will be... $mathbb{Z}$.
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
answered Dec 22 '18 at 10:31
A.RodA.Rod
57935
57935
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
1
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
add a comment |
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
1
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
$begingroup$
Thanks, I have found out where I went wrong. I tacitly assumed that every point not in $X$ had an open neighbourhood disjoint from $X$. But for the points in $X$, the stalks are the same, aren't they?
$endgroup$
– Jehu314
Dec 22 '18 at 13:06
1
1
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
$begingroup$
Yes, by definition of the induced topology.
$endgroup$
– A.Rod
Dec 22 '18 at 13:09
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049243%2fstalks-of-the-pushforward-sheaf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
