How to construct $X \oplus \Sigma X$ from $X \oplus \Sigma X \oplus \Sigma X \oplus \Sigma^2 X$ without splitting an idempotent? 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?
Notes:

*

*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.


*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.


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