TITLE: X is a direct summand of $X\otimes DX\otimes X$ in a rigid tensor-triangulated category.
QUESTION [1 upvotes]: I am reading Balmer's "Supports and filtrations in algebraic geometry and modular representation theory". In Prop. 2.4 he claims that $X$ (an object of a rigid tensor triangulated category) is a direct summand of $X\otimes DX\otimes X$ (see this topic All Ideals are Radical in Rigid Categories). However, my impression is that the proof only shows that $X$ is a rectract of $X\otimes DX\otimes X$. It looks like the distinction between those two notions is quite crucial for the later applications (thickness of some ideals). Later in this paper he assumes additionally that the category is also idempotent complete, in which case retract happens to be the same as a direct summand.
Coproducts are distributive under the tensor product (since they are adjoints), so one could try to show that $1$ is a direct summand of $X\otimes DX$, but I was unable to show that.
Any help would be highly appreciated.

REPLY [5 votes]: In a triangulated category a retract is a direct summand.
Suppose $X$ is a retract of $Y$. So there are maps $\alpha: X\to Y$ and $\beta: Y\to X$ with $\beta\alpha=\text{id}_X$. Complete $\alpha$ to a distinguished triangle
$$Z[-1]\xrightarrow{\gamma}X\xrightarrow{\alpha}Y\rightarrow Z.$$
Then $\gamma=\beta\alpha\gamma=0$, so the triangle is isomorphic to
$$Z[-1]\xrightarrow{0}X\rightarrow X\oplus Z\rightarrow Z.$$
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