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According to the Elephant, and these notes, 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, then X is indecomposable iff it is connected (proof and details here).

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

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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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.

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 lextensive 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 $[\mathbf{C}^\mathrm{op},\mathbf{Set}]$.)

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Thanks, this is exactly the sort of thing I was looking for. –  Finn Lawler Aug 17 '10 at 16:56
    
Thanks, glad to help! Sorry I can't suggest a reference though… –  Peter LeFanu Lumsdaine Aug 17 '10 at 18:36
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Here's a good reference for this sort of thing: tiny.cc/pvgq6 –  Andrej Bauer Aug 17 '10 at 23:17
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Already in a pretopos all epis are regular; see A1.4.9 in the Elephant. However, even in a topos the Elephant definition does not imply the LS definition except for projective objects. For instance, if X is a connected space which is the (non-disjoint) union of two proper open subsets U and V, then in the topos Sh(X) the terminal object is not a nontrivial coproduct, but the map from U+V to the terminal object is epi although neither component is so. –  Mike Shulman Aug 18 '10 at 4:50
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I suppose computer jokes don't work on a math forum... –  Andrej Bauer Aug 18 '10 at 7:49
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