An object $X$ of a category $C$ is **indecomposable** if it cannot be expressed as a non-trivial coproduct of objects of $C$. Formally, $X$ is indecomposable if given an isomorphism $X \cong \coprod_i U_i$, there is a unique index $i$ such that $X \cong U_i$ and $U_j \cong 0$ for all $j \neq i$, where $0$ is an initial object.

The requirement that $i$ be unique keeps the initial object itself from being indecomposable; this is analogous to being too simple to be simple.

If $C$ is an extensive category, meaning that coproducts in $C$ are disjoint and pullback-stable, then we have

An object $X$ of $C$ is indecomposable if and only if it is connected, that is if the hom functor $\hom(X,-)$ preserves coproducts.

If $X$ is connected, then a morphism $k \colon X \to \coprod_i U_i$ factors uniquely as $\iota_i \circ \bar k \colon X \to U_i \to \coprod_i U_i$, where $\iota_i$ is the coprojection. Suppose $k$ is invertible – then the composite $k^{-1} \iota_i \bar k \colon X \to U_i \to \coprod_i U_i \to X$ is the identity. Consider

$U_i \overset{\iota_i}{\to} \coprod_i U_i \overset{k^{-1}}{\to} X \overset{k}{\to} \coprod_i U_i$

Of course $k \circ k^{-1} = 1$, while $k = \iota_i \bar k$ as before. So $\iota_i \circ \bar k k^{-1} \iota_i = \iota_i$. But because $C$ is extensive the coprojections are monic, so $\bar k k^{-1} \iota_i = 1$. Thus $\bar k$ is an isomorphism $X \cong U_i$, with inverse $k^{-1} \iota_i$. Because coproducts in $C$ are disjoint, the pullback of distinct coprojections is $0$, and because $U_i \cong X \cong \coprod_i U_i$, the pullback of $\iota_i$ along $\iota_j$ is an isomorphism, showing that $U_j \cong 0$ for $j \neq i$.

Conversely, assume $X$ is indecomposable. Given $k \colon X \to \coprod_i U_i$, we are to produce a unique $X \to U_i$ as above. Because $C$ is extensive, $k$ is isomorphic (in the slice category $C/\coprod_i U_i$) to $\coprod_i k_i \colon \coprod_i X_i \to \coprod_i U_i$ for some family $\{k_i \colon X_i \to U_i\}$. But $X \cong \coprod_i X_i$ gives us by indecomposability of $X$ an isomorphism $X \cong X_i$ for a unique $i$, which composed with $k_i$ gives a morphism $X \cong X_i \to U_i$. Because $\coprod_i k_i = [\iota_i k_i]_i$ (where the right-hand side is the copairing of the family $\{\iota_i k_i\}_i$), and because the $X_j \cong 0$ for $j\neq i$, the composite $\iota_i k_i \colon X \cong X_i \to U_i \to \coprod_i U_i$ is equal to $k$. Hence $X$ is connected.

If $C$ is a presheaf category $[S^{op}, Set]$ (thus a Grothendieck topos and so *a fortiori* (infinitary) extensive), then it is easy to see that the representable functors $S(-,s)$ are connected and so indecomposable. Conversely, the objects of $[S^{op}, Set]$ that are indecomposable as well as projective are precisely the objects of the Cauchy completion of $S$.

Lambek Scott give a different definition of indecomposability. Generalizing their definition slightly, we may say that an object $X$ is **indecomposable** (in the sense of Lambek–Scott) if any jointly epimorphic family $\{U_i \to X\}_i$ of arrows into $X$ contains at least one epimorphism $U_i \twoheadrightarrow X$, and moreover the unique arrow $0 \to X$ is not epic (this to ensure that $0$ is not indecomposable).

If the epi $U_i \twoheadrightarrow X$ is required to be regular, then in an *extensive category* the Lambek–Scott definition implies that given above: if $k \colon X \cong \coprod_i U_i$, then the family $\{k^{-1} \iota_i \colon U_i \to \coprod_i U_i \cong X\}_i$ is jointly epic, so it contains a regular epi $\iota_i k^{-1}$. But extensivity implies that $\iota_i$ is a monomorphism, so the regular epi $\iota_i k^{-1}$ is also monic and hence an isomorphism. The converse does not hold in general, but it does hold if $X$ is projective. See this MathOverflow thread for a discussion.

An indecomposable representation is precisely an indecomposable object in an appropriate category $Rep$ of representations, as one would expect. In contrast, an irreducible representation is precisely a simple object in $Rep$. Every irreducible representation is indecomposable, but the converse holds only in special situations (such as the category of finite-dimensional linear representations of a real semisimple Lie group).

However, one level decategorified, an irreducible element of a poset $P$ is precisely an indecomposable object of $P$ when thought of as a thin category. In contrast, a simple object is analogous to an atomic element, although they are not the same thing. (One might say that atomic = $0$-simple.) Again, every atomic element is irreducible, but the converse holds only in special situations (such as the power set of any set).

The bottom line is that ‘irreducible’ and ‘indecomposable’ sometimes mean the same thing but sometimes don't, and ‘irreducible’ doesn't even mean the same thing across different fields.

Last revised on September 20, 2011 at 00:46:42. See the history of this page for a list of all contributions to it.