In truth, I do not need in the general case.


Let $R$ be a ring (associative, with unit element) and let $\mathrm{Comp}(R)$ the *category of complexes* over $\mathrm{Mod}\ R$.


If $\mathbb{N}$ is the natural numbers with the usual partial order and $\{X\_i^\bullet, \varphi\_{i,j}\colon X\_j^\bullet \to X\_i^\bullet\}$ is an *inverse system* over $\mathbb{N}$ such that $X\_i^\bullet \in \mathrm{Comp}(R)$ is indecomposable, for all $i \in \mathbb{N}$, so is indecomposable $\varprojlim X\_i^\bullet$?


**Edit.** Of course, I'm assuming $\varphi\_{i,j} \neq 0$ for all $j\geq i \in \mathbb{N}$.




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**Remark.** A category $\mathcal{C}$ is *complete* if $\varprojlim A\_i$ exists in $\mathcal{C}$ for every inverse system $\{A\_i , \varphi\_{i,j}\}$ in $\mathcal{C}$. The categories $\mathcal{C} = \mathrm{Mod} \ R$ and $\mathcal{C} = \mathrm{Comp}(R)$ are examples of complete categories.



Is a inverse limit of indecomposable again indecomposable?
Suppose $M$ and $M'$ are two $R$-modules with a common submodule $N$ that has a descending chain of submodules
$$N=N\_0\supseteq N\_1\supseteq N\_2\supseteq\dots$$
with zero intersection. 


Then if $X^\bullet\_i$ is the two term complex
$$\dots\to 0\to N\_i\to M\oplus M'\to 0\to\dots,$$
where the differential embeds $N\_i$ diagonally into $M\oplus M'$, the inverse limit of the $X^\bullet\_i$ is the complex
$$\dots\to 0\to 0\to M\oplus M'\to 0\to\dots,$$
which is decomposable, but it's not too hard to construct cases where all the complexes $X^\bullet\_i$ are indecomposable.


For example, for complexes of abelian groups, take $M$ to be the $p$-adic integers $\mathbb{Z}\_p$, and $M'=\mathbb{Z}\_q$ for primes $p\neq q$, and take $N=\mathbb{Z}$ with the chain of subgroups $N\_i=2^i\mathbb{Z}$.


