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\begin{document}

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\section*{indecomposable object}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{category_theory}{}\paragraph*{{Category theory}}\label{category_theory}

[[!include category theory - contents]]

\hypertarget{topos_theory}{}\paragraph*{{Topos Theory}}\label{topos_theory}

[[!include topos theory - contents]]

\hypertarget{indecomposable_objects}{}\section*{{Indecomposable objects}}\label{indecomposable_objects}

\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{remarks}{Remarks}\dotfill \pageref*{remarks} \linebreak
\noindent\hyperlink{lambekscott_indecomposability}{Lambek--Scott indecomposability}\dotfill \pageref*{lambekscott_indecomposability} \linebreak
\noindent\hyperlink{irreducible}{Indecomposability vs irreducibility}\dotfill \pageref*{irreducible} \linebreak


\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

An object $X$ of a category $C$ is \textbf{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]].

\hypertarget{remarks}{}\subsection*{{Remarks}}\label{remarks}

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

\begin{uprop}
An object $X$ of $C$ is indecomposable if and only if it is [[connected object|connected]], that is if the hom functor $\hom(X,-)$ preserves coproducts.

\end{uprop}
\begin{proof}
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

\begin{displaymath}
U_i \overset{\iota_i}{\to} \coprod_i U_i \overset{k^{-1}}{\to} X \overset{k}{\to} \coprod_i U_i
\end{displaymath}
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.

\end{proof}
If $C$ is a [[presheaf category]] $[S^{op}, Set]$ (thus a [[Grothendieck topos]] and so \emph{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 object|projective]] are precisely the objects of the [[Cauchy completion]] of $S$.

\hypertarget{lambekscott_indecomposability}{}\subsection*{{Lambek--Scott indecomposability}}\label{lambekscott_indecomposability}

[[Introduction to Higher-Order Categorical Logic|Lambek \& Scott]] give a different definition of indecomposability. Generalizing their definition slightly, we may say that an object $X$ is \textbf{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 epimorphism|regular]], then in an \emph{[[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 object|projective]]. See \href{http://mathoverflow.net/questions/35855/indecomposable-objects-in-a-category}{this MathOverflow thread} for a discussion.

\hypertarget{irreducible}{}\subsection*{{Indecomposability vs irreducibility}}\label{irreducible}

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.

[[!redirects indecomposable object]] [[!redirects indecomposable objects]]



\end{document}