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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.

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$$(X\oplus Y)\oplus(X\oplus Y)\oplus(X\oplus Y)\oplus\dots\cong X\oplus (Y\oplus X)\oplus(Y\oplus X)\oplus\dots,$$ so there is a triangle $$X\to(X\oplus Y)^{(\mathbb{N})}\to(X\oplus Y)^{(\mathbb{N})}\to\Sigma X,$$ where $(X\oplus Y)^{(\mathbb{N})}$ denotes the coproduct of countably many copies of $X\oplus Y$.

This is often called the "Eilenberg swindle".

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  • $\begingroup$ 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. $\endgroup$ Dec 17 '18 at 13:13
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Using the language of $\alpha$-localising subcategories (Take a triangulated category $T$ with arbitrary coproducts and let $\alpha$ be a regular cardinal, $\alpha$-localising subcategories are closed under coproducts of fewer than $\alpha$ objects), I believe it is said that $\alpha$-localising subcategories are thick if $\alpha > \aleph_0$.

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