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](https://doi.org/10.1023/A:1017932514274)) 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:**


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.



How to construct $X \oplus \Sigma X$ from $X \oplus \Sigma X \oplus \Sigma X \oplus \Sigma^2 X$ without splitting an idempotent?
The cofibre of $0\oplus 1\oplus 1\oplus 1$ on $X=Z\oplus \Sigma Z\oplus\Sigma Z\oplus\Sigma^2Z$ is $Z\oplus\Sigma Z=Y$.


