Are localizing subcategories thick?
$begingroup$
We work in $T$ triangulated category admitting small coproducts, we say $S$ a full subcategory is exact or triangulated iff it is closed under suspensions and triangles. Moreover such $S$ is called localising whenever it is closed under infinite coproducts and thick if it is closed under summands, i.e. if $X oplus Y in S$ then both $X$ and $Y$ are in $S$.
These are classical definitions in the theory of triangulated category which I think are familiar to anyone who studied it.
I was told that every localizing subcategory is thick but I cannot see this from the definitions.
My first thought was using the triangle $X rightarrow X oplus Y rightarrow Y rightarrow Sigma X$ to get the claim but I only know $X oplus Y in S$ so I cannot conclude using the closure under triangles.
Does this follows from the definitions alone or do I need some result? My intuition tells me that if the claim is true it should be an easy fact which does not require a complicate proof.
Maybe I need an additional assumption on $T$?
Edit: I recalled now that if $T$ is compactly generated then Brown representability theorem implies the existence of a localization functor $L colon T rightarrow T$ with kernel $S$ and such $L$ is left adjoint to the inclusion $S^perp rightarrow T$. Thus $L(X oplus Y)cong LX oplus LY$ and this is zero iff both $LX$ and $LY$ are zero. This gives the claim I wanted but the proof is more complex that what I thought at first and I need the additional assumption that $T$ must be compactly generated. Can you provide an easier proof or one which does not require the compactly generated hypothesis? Or give a counterexample if such exists.
triangulated-categories
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
$endgroup$
add a comment |
$begingroup$
We work in $T$ triangulated category admitting small coproducts, we say $S$ a full subcategory is exact or triangulated iff it is closed under suspensions and triangles. Moreover such $S$ is called localising whenever it is closed under infinite coproducts and thick if it is closed under summands, i.e. if $X oplus Y in S$ then both $X$ and $Y$ are in $S$.
These are classical definitions in the theory of triangulated category which I think are familiar to anyone who studied it.
I was told that every localizing subcategory is thick but I cannot see this from the definitions.
My first thought was using the triangle $X rightarrow X oplus Y rightarrow Y rightarrow Sigma X$ to get the claim but I only know $X oplus Y in S$ so I cannot conclude using the closure under triangles.
Does this follows from the definitions alone or do I need some result? My intuition tells me that if the claim is true it should be an easy fact which does not require a complicate proof.
Maybe I need an additional assumption on $T$?
Edit: I recalled now that if $T$ is compactly generated then Brown representability theorem implies the existence of a localization functor $L colon T rightarrow T$ with kernel $S$ and such $L$ is left adjoint to the inclusion $S^perp rightarrow T$. Thus $L(X oplus Y)cong LX oplus LY$ and this is zero iff both $LX$ and $LY$ are zero. This gives the claim I wanted but the proof is more complex that what I thought at first and I need the additional assumption that $T$ must be compactly generated. Can you provide an easier proof or one which does not require the compactly generated hypothesis? Or give a counterexample if such exists.
triangulated-categories
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
$endgroup$
add a comment |
$begingroup$
We work in $T$ triangulated category admitting small coproducts, we say $S$ a full subcategory is exact or triangulated iff it is closed under suspensions and triangles. Moreover such $S$ is called localising whenever it is closed under infinite coproducts and thick if it is closed under summands, i.e. if $X oplus Y in S$ then both $X$ and $Y$ are in $S$.
These are classical definitions in the theory of triangulated category which I think are familiar to anyone who studied it.
I was told that every localizing subcategory is thick but I cannot see this from the definitions.
My first thought was using the triangle $X rightarrow X oplus Y rightarrow Y rightarrow Sigma X$ to get the claim but I only know $X oplus Y in S$ so I cannot conclude using the closure under triangles.
Does this follows from the definitions alone or do I need some result? My intuition tells me that if the claim is true it should be an easy fact which does not require a complicate proof.
Maybe I need an additional assumption on $T$?
Edit: I recalled now that if $T$ is compactly generated then Brown representability theorem implies the existence of a localization functor $L colon T rightarrow T$ with kernel $S$ and such $L$ is left adjoint to the inclusion $S^perp rightarrow T$. Thus $L(X oplus Y)cong LX oplus LY$ and this is zero iff both $LX$ and $LY$ are zero. This gives the claim I wanted but the proof is more complex that what I thought at first and I need the additional assumption that $T$ must be compactly generated. Can you provide an easier proof or one which does not require the compactly generated hypothesis? Or give a counterexample if such exists.
triangulated-categories
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
$endgroup$
We work in $T$ triangulated category admitting small coproducts, we say $S$ a full subcategory is exact or triangulated iff it is closed under suspensions and triangles. Moreover such $S$ is called localising whenever it is closed under infinite coproducts and thick if it is closed under summands, i.e. if $X oplus Y in S$ then both $X$ and $Y$ are in $S$.
These are classical definitions in the theory of triangulated category which I think are familiar to anyone who studied it.
I was told that every localizing subcategory is thick but I cannot see this from the definitions.
My first thought was using the triangle $X rightarrow X oplus Y rightarrow Y rightarrow Sigma X$ to get the claim but I only know $X oplus Y in S$ so I cannot conclude using the closure under triangles.
Does this follows from the definitions alone or do I need some result? My intuition tells me that if the claim is true it should be an easy fact which does not require a complicate proof.
Maybe I need an additional assumption on $T$?
Edit: I recalled now that if $T$ is compactly generated then Brown representability theorem implies the existence of a localization functor $L colon T rightarrow T$ with kernel $S$ and such $L$ is left adjoint to the inclusion $S^perp rightarrow T$. Thus $L(X oplus Y)cong LX oplus LY$ and this is zero iff both $LX$ and $LY$ are zero. This gives the claim I wanted but the proof is more complex that what I thought at first and I need the additional assumption that $T$ must be compactly generated. Can you provide an easier proof or one which does not require the compactly generated hypothesis? Or give a counterexample if such exists.
triangulated-categories
triangulated-categories
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
edited Dec 14 '18 at 17:36
N.B.
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
asked Dec 14 '18 at 17:25
N.B.N.B.
696313
696313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
so there is a triangle
$$Xto(Xoplus Y)^{(mathbb{N})}to(Xoplus Y)^{(mathbb{N})}toSigma X,$$
where $(Xoplus Y)^{(mathbb{N})}$ denotes the coproduct of countably many copies of $Xoplus Y$.
This is often called the "Eilenberg swindle".
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
$endgroup$
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
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%2f3039671%2fare-localizing-subcategories-thick%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
$begingroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
so there is a triangle
$$Xto(Xoplus Y)^{(mathbb{N})}to(Xoplus Y)^{(mathbb{N})}toSigma X,$$
where $(Xoplus Y)^{(mathbb{N})}$ denotes the coproduct of countably many copies of $Xoplus Y$.
This is often called the "Eilenberg swindle".
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
$endgroup$
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
add a comment |
$begingroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
so there is a triangle
$$Xto(Xoplus Y)^{(mathbb{N})}to(Xoplus Y)^{(mathbb{N})}toSigma X,$$
where $(Xoplus Y)^{(mathbb{N})}$ denotes the coproduct of countably many copies of $Xoplus Y$.
This is often called the "Eilenberg swindle".
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
$endgroup$
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
add a comment |
$begingroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
so there is a triangle
$$Xto(Xoplus Y)^{(mathbb{N})}to(Xoplus Y)^{(mathbb{N})}toSigma X,$$
where $(Xoplus Y)^{(mathbb{N})}$ denotes the coproduct of countably many copies of $Xoplus Y$.
This is often called the "Eilenberg swindle".
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
$endgroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
so there is a triangle
$$Xto(Xoplus Y)^{(mathbb{N})}to(Xoplus Y)^{(mathbb{N})}toSigma X,$$
where $(Xoplus Y)^{(mathbb{N})}$ denotes the coproduct of countably many copies of $Xoplus Y$.
This is often called the "Eilenberg swindle".
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
answered Dec 16 '18 at 12:19
Jeremy RickardJeremy Rickard
16.4k11743
16.4k11743
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
add a comment |
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
The second map in the triangle is the one twisting the sum $X oplus Y$ in $Y oplus X$ and then injecting infinitely many such copies in $X oplus (Y oplus X) oplus (Y oplus X) dots$, right?
$endgroup$
– N.B.
Dec 17 '18 at 9:24
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
$begingroup$
@N.B. Yes, right.
$endgroup$
– Jeremy Rickard
Dec 17 '18 at 13:13
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%2f3039671%2fare-localizing-subcategories-thick%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
