According to the Elephant, and [these notes](http://www.staff.science.uu.nl/~ooste110/syllabi/toposmoeder.pdf), an object X in a category C is *indecomposable* if given an isomorphism $X \cong \coprod\_i U\_i$ there is a unique $i$ such that $X \cong U\_i$ and $U\_j \cong 0$ for $j\neq i$ where 0 is the initial object. If C is [extensive](http://ncatlab.org/nlab/show/extensive+category), then X is indecomposable iff it is [connected](http://ncatlab.org/nlab/show/connected+object) (proof and details [here](http://ncatlab.org/nlab/show/indecomposable+object)).


Lambek and Scott give a different definition: they say that X is indecomposable if given an epi $[k,l] \colon U + V \twoheadrightarrow X$, one of k or l is epi. I suppose this can be generalised to say that X is indecomposable if any jointly epic family into X contains at least one epi.


Perhaps I'm missing something obvious, but I can't see that either definition implies the other. So my question is



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> Are these definitions equivalent, or does one imply the other, in general or in some specific class of categories? Do you know of a reference that compares or discusses the two?
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Indecomposable objects in a category

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> **Briefly:** there's a simple difference in how they treat 0. That fixed, still neither implies the other in general. In a regular extensive category, a slight modification of the LS definition implies the Elephant one. I suspect they're not fully equivalent in anything short of a topos. As Mike Shulman points out, even in a topos they are not equivalent.
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The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty). So, let's temporarily change one of the definitions to fix this. I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition. (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)


In eg **Top**, however, we can see that the Elephant def still doesn't imply the LS def. $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi). Even more, it’s decomposable by a *regular* epi (more on this distinction below).


Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but *both* of $f$ and $g$ are mono).


When *do* they imply each other? If we upgrade the LS definition to involve *regular* epis, then in a [regular](http://ncatlab.org/nlab/show/regular+category) [lextensive](http://ncatlab.org/nlab/show/extensive+category) category, it implies the Elephant definition, if I'm not mistaken. For this, suppose $X$ is “indecomposable by reg epis”, and suppose $X \cong A + B$ — WLOG $X = A + B$. The coproduct inclusions are then jointly reg epi, so one of them is reg epi. But it's also mono (in a lextensive category, every coproduct inclusion is a pullback of $1 \to 1 + 1$, so is mono); so it's iso. There's a little more fiddly stuff to check involving messing around with $0$, but it's all the same sort of thing.


**Edit from Mike Shulman's comments:** if moreover we're in a pretopos, all epis are regular, so there the original LS definition will imply the Elephant definition. On the other hand, the Elephant definition doesn't imply the LS even in a topos: the terminal object of $\mathbf{Sh}([0,1])$ is a counterexample, essentially for the same reasons that $[0,1]$ was a counterexample in $\mathbf{Top}$.


However, the two definitions are equivalent for *projective* objects… and I guess that's how this situation has arisen, since a common use of indecomposable objects in topos theory is the theorem that the indecomposable projectives in a presheaf category are exactly the retracts of representables. (This is useful because it lets us recover the idempotent-completion of $\mathbf{C}$, which is very close to $\mathbf{C}$ itself, from $[\mathbf{C}^\mathrm{op},\mathbf{Set}]$.)


