Let $P^{\bullet} = (P^i, d^i)\_{i\leq 0}$ be an **indecomposable** object in category of complexes bounded above, where each $P\_i$ is a finitely generated projective module over an finite dimensional algebra. Denote by $\tau^{\geq -n} (P^{\bullet})$ the brutal truncation of $P^{\bullet}$, $n \geq 1$, i.e.,
$$
\tau^{\geq -n} (P^{\bullet}) = \cdots \to 0 \to P^{-n} \to \cdots \to P^{-1} \to P^{0} \to 0 \to \cdots
$$ 


Is it **indecomposable** $\tau^{\geq -n} (P^{\bullet})$ for all $n \geq 1$?


Thanks!


**Edit**: 


– Suppose $P^{\bullet}$ is **unbounded below**, i.e., $P^{\bullet} \in \mathcal{C}^{-}(\text{proj}-A) \setminus \mathcal{C}^{b}(\text{proj}-A)$.


– Suppose $P^0 \neq 0$, in order to fix notation. 



Brutal truncation of indecomposable complexes
No. Let $k$ be a field, and let $A = k[x,y]/(x^2, y^2, xy, yx)$. Then the morphism $A \to A\oplus A$ sending $1$ to $(x,y)$ gives an indecomposable complex
$$P^\bullet: \ldots \to 0 \to A\to A\oplus A\to 0 \to \ldots $$
If you put $A\oplus A$ in degree $-1$, then $\tau^{\geq -1}(P^\bullet)$ is not indecomposable. Notice that the image of the map described above lies in the radical of $A\oplus A$.




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**Edit:** An example which is unbounded below can be obtained using the same algebra $A$. Take the following complex:
$$ Q^\bullet : \ldots \longrightarrow A\oplus A \stackrel{\begin{pmatrix}y & 0 \\ x & 0\end{pmatrix}}{\longrightarrow} A\oplus A \stackrel{\begin{pmatrix}y & 0 \\ x & 0\end{pmatrix}}{\longrightarrow} A\oplus A \stackrel{\begin{pmatrix}x & 0 \end{pmatrix}}{\longrightarrow} A \longrightarrow 0 \longrightarrow \ldots,$$
where $A$ is in degree $0$. Then $Q^\bullet$ is indecomposable, but $\tau^{\geq -n}(Q^\bullet)$ has a direct summand given by the complex with $A$ concentrated in degree $-n$, so it is not indecomposable.


