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.
701313
$endgroup$
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.
701313
$endgroup$
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.
701313
$endgroup$
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.
701313
asked Dec 14 '18 at 17:25
N.B.N.B.
701313
edited Dec 14 '18 at 17:36
N.B.
asked Dec 14 '18 at 17:25
N.B.N.B.
701313
asked Dec 14 '18 at 17:25
N.B.N.B.
701313
asked Dec 14 '18 at 17:25
N.B.N.B.
701313
701313
add a comment |
add a comment |
1 Answer
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$begingroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
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 |
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$begingroup$
$$(Xoplus Y)oplus(Xoplus Y)oplus(Xoplus Y)oplusdotscong Xoplus (Yoplus X)oplus(Yoplus X)oplusdots,$$
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 |
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