Let $Z$ be an object in a stable (or triangulated/whatever) category $\mathcal C$. I believe it follows from Thomason's theorem (see The classification of triangulated subcategories) that the triangulated categories generated by $$Y = Z \oplus \Sigma Z$$ and $$X = Y \oplus \Sigma Y = Z \oplus \Sigma Z \oplus \Sigma Z \oplus \Sigma^2 Z$$ are the same. This means that $X$ and $Y$ can each be constructed from one one another in finitely many steps, where each step consists of taking the fiber or cofiber of a map between previously-constructed objects. One direction of this is obvious — clearly $X$ can be constructed from $Y$ in this way. The other direction is not so clear to me.

Question: Let $Z$ be an object in a stable category $\mathcal C$, and let $Y,X$ be as above. How can one explicitly construct $Y$ from $X$ in finitely many steps, using just fibers and cofibers?


  1. It takes finitely many steps to construct a de/suspension, extension, or direct sum from fibers and cofibers, so one is free to use these as steps as well.

  2. The argument from Thomason's theorem doesn't actually rely on $Y$ being of the form $X \oplus \Sigma X$; it's only important that $Y$ represent the zero class in the $K$-theory $G = K_0(\langle Y \rangle_{thick})$ of the thick subcategory it generates. The argument then is that $Y$ and $Y \oplus \Sigma Y$ both generate the zero subgroup of $G$, and both generate $\langle Y \rangle_{thick}$ as thick subcategories (they are "dense" in Thomason's sense) and so by Thomason's theorem, the triangulated categories they generate are the same.

  3. In light of (2), it may be better simply to assume that $Y$ represents the zero class in $K$-theory. On the other hand, it's possible that the construction I'm looking for will be more explicit in the case where $Y = X \oplus \Sigma X$, and if so, then I'm interested in having this extra explicit-ness.


2 Answers 2


The cofibre of $0\oplus 1\oplus 1\oplus 1$ on $X=Z\oplus \Sigma Z\oplus\Sigma Z\oplus\Sigma^2Z$ is $Z\oplus\Sigma Z=Y$.

  • $\begingroup$ Great, thanks! I should have seen that one. $\endgroup$
    – Tim Campion
    Sep 22, 2022 at 15:12

Whenever $P$ is a summand of $Q$, you can construct $P\oplus\Sigma P$ in one step from $Q$: if $e$ is the idempotent that projects onto $P$, then the cofiber of $1-e$ is $P\oplus\Sigma P$.

You can apply this to the summand $Z$ of $X$.

There are obstructions to the more general case when $Y$ is $0$ in $K_0$ : for any $Y$ in any stable $\infty$-category $C$, embed $C$ in a bigger one whose $K_0$ vanishes. The question of whether $Y\in \langle Y\oplus\Sigma Y\rangle$ does not depend on whether you are in $C$ or this bigger one, but the vanishing of the class of $Y$ in $K_0$ does.

Basically, by Thomason's result it boils down to the class of $Y$ in $K_0(\langle Y \rangle)$. If it vanishes, then by Thomason's result you get what you want, and if it doesn't, that's an obstruction. The vanishing of $Y$ in $K_0(\langle Y\rangle )$ means, taking $R = End(Y)$, that the class of $R$ is zero in $K_0(Perf(R))$.

Say for a second that $R$ is connective, then this means (using the fact that we can compute this $K_0$ via projectives) that there exists a finite $n$ with $R^n\simeq R\oplus R^n$, $R$-linearly and from this it is easy to construct $R$ from $R\oplus \Sigma R$.

Here's one construction: observe that it follows that $R^n\oplus R^n\simeq R^n$, then construct $R^n\oplus\Sigma R^n$ as a sum of $R\oplus \Sigma R$'s, and then from that $R^n\oplus R^n \oplus \Sigma R^n$ , and from that together with $R^n\oplus\Sigma R^n$, you get $R^n$ as a cofiber, and then you can chop that off $R\oplus R^n$ (which you have, because it's $R^n$ !) to get $R$.

In general, a witness for the $0$-ness of the class of $R$ will give you a construction of $R$ from $R\oplus \Sigma R$. Here's how it works : because $K_0$ of your stable category is the quotient of its additive $K_0$ by the obvious relations imposed by cofiber sequences, the assumption that $[R] = 0$ gives you (that's an easy exercise) a cofiber sequence $A\to B\to C$ of $R$-modules such that $R\oplus A\simeq B\oplus \Omega C$. Let me call $\eta: C\to \Sigma A$ the corresponding map.

From $R\oplus\Sigma R$ you can construct $M\oplus\Sigma M$ for any perfect module $M$, and in particular $(C\oplus A)\oplus (\Sigma C\oplus \Sigma A)$. This module has a self map given by the following : $C$ maps to $\Sigma A$ via $\eta$, $\Sigma A$ maps to $0$, and the other summands map to themselves via the identity. The fiber of this self map is $B\oplus \Sigma A\oplus \Omega C\simeq R\oplus A\oplus\Sigma A$.

But you can also construct $A\oplus\Sigma A$ from $R\oplus \Sigma R$, so you can mod it out and you have constructed $R$.

  • $\begingroup$ Thanks -- this is great, especially as it addresses the broader context. If I could accept an additional answer I definitely would! $\endgroup$
    – Tim Campion
    Sep 22, 2022 at 15:20
  • $\begingroup$ I've added a "constuction" in the general case. Basically from a witness of $[R] = 0$ you can construct $R$ from $R\oplus\Sigma R$. $\endgroup$ Sep 28, 2022 at 13:57

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