A category is exhaustive if it has colimits of (transfinite) sequences of monomorphisms which interact well with pullbacks.
Given a transfinite sequence
(continuing through $A_\alpha$ for all $\alpha\lt\kappa$, where $\kappa$ is some limit ordinal) for which each transition map $A_\alpha \to A_\beta$ is a monomorphism, if $A_\kappa$ is a colimit of this sequence (a transfinite composition) we call it a transfinite union.
A category $C$ is exhaustive if one, hence both, of the following two equivalent conditions hold.
$C$ has transfinite unions which are stable under pullback and for which the coprojections $A_\alpha \to A_\kappa$ are also monomorphisms.
$C$ has transfinite unions, and given any diagram
in which the bottom row is a transfinite union and for each $\alpha\lt\beta\lt\kappa$, the square
is a pullback (hence $B_\alpha \to B_\beta$ is also monic), then the top row is a transfinite union (i.e. a colimit diagram) if and only if for all $\alpha\lt\kappa$ the square
is a pullback. (In other words, transfinite unions are van Kampen colimits.)
Note that half of the second condition is simply that the colimits in the first condition are stable under pullback. One may obtain various weaker notions by restricting the allowable values of $\kappa$.
We will prove the equivalence of these two characterizations in Propositions and below.
The category Set is easily verified to be exhaustive.
Since limits and colimits in presheaf categories are pointwise, any presheaf category $[D^{op},Set]$ is also exhaustive.
Since the condition involves only colimits and finite limits, it is preserved by any left exact localization. Thus, every Grothendieck topos is exhaustive.
We first prove the equivalence of the above two definitions.
In a category which satisfies the second definition above, given a transfinite union
(so that $A_\alpha \to A_\beta$ is monic for all $\alpha\lt\beta\lt\kappa$), the coprojections $A_\alpha\to A_\kappa$ are also monic.
For a fixed $\alpha$, consider the diagram
It is easy to see that each intermediate square is a pullback, and of course the top row is a colimit. Thus, in particular the square
is a pullback, i.e. $A_\alpha\to A_\kappa$ is monic.
Suppose $C$ satisfies the first condition above, i.e. it has pullback-stable transfinite unions for which the coprojections are also monic. Then $C$ also satisfies the second condition above.
It remains to show that given a diagram
in which both rows are transfinite compositions of monomorphisms and the squares
are pullbacks, also the squares
are pullbacks.
Fix an $\alpha\lt\kappa$, and suppose given $g\colon X\to B_\kappa$ such that $f_\kappa g$ factors through $A_\alpha$; we want to show that $g$ factors through $B_\alpha$. Pulling back the $B$-colimit along $g$ and the $A$-colimit along $f_\kappa g$, we obtain another morphism of transfinite unions:
and it is easy to verify, by the pasting law for pullbacks, that each square
is a pullback for $\beta\lt\gamma\lt\kappa$. However, since $f_\kappa g$ factors through $A_\alpha$, the morphism $(f_\kappa g)^* A_\beta \to (f_\kappa g)^* A_\gamma$ is split epic whenever $\alpha\lt\beta\lt\gamma$, hence (since it is also monic) an isomorphism. Thus, in this case so is its pullback $g^* B_\beta \to g^* B_\gamma$. Since the coprojections into a transfinite composite of a sequence that eventually consists of isomorphisms are also eventually isomorphisms, $g^* B_\alpha \to X$ is an isomorphism; hence $g$ factors through $B_\alpha$ as desired.
Exhaustiveness also interacts well with other exactness properties:
If an exhaustive category is also finitary extensive, then it is infinitary extensive.
A coproduct of cardinality $\kappa$ can be constructed using transfinite compositions and finite coproducts:
Finitary extensivity implies that all the transition maps are monic, so that these transfinite composites exist in an exhaustive category. Pullback-stability of coproducts follows easily from pullback-stability of transfinite unions and of finite coproducts. Finally, it is known (see extensive category) that if finite coproducts exist and are disjoint, then any infinite coproducts which exist are also disjoint.
Colimits in an exhaustive category preserve finite limits.
In an exhaustive category, if $X$ is the $\kappa$-indexed transfinite union of a sequence $(A_\alpha)_{\alpha\lt\kappa}$ and also a sequence $(B_\alpha)_{\alpha\lt\kappa}$, then it is also the transfinite union of the sequence $(A_\alpha\cap B_\alpha)_{\alpha\lt\kappa}$ (the intersection being taken as subobjects of $X$).
Since transfinite unions are stable under pullback, $X$ is the colimit of the $(\kappa\times\kappa)$-indexed diagram $(A_\alpha\cap B_\beta)_{\alpha\lt\kappa, \beta\lt\kappa}$. But $\kappa$ is a filtered category, so its diagonal functor is final.
In an exhaustive category, given a pullback diagram of $\kappa$-indexed transfinite sequences of monomorphisms:
the induced square of colimits:
is also a pullback.
Define $P$ so that the following square is a pullback:
Since pullbacks preserve transfinite unions, $P$ is the transfinite union of the sequnece $(f^* B_\alpha)_{\alpha\lt\kappa}$ and also the sequence $(g^* C_\alpha)_{\alpha\lt\kappa}$. Thus, by Lemma , $P$ is also the transfinite union of $(f^* B_\alpha\cap g^* C_\alpha)_{\alpha\lt\kappa}$. However, using the fact that the inclusions $D_\alpha \to D_\kappa$ are monic, it is straightforward to verify that
Thus, $P$ is the transfinite union of $(A_\alpha)_{\alpha\lt\kappa}$, hence $P\cong A_\kappa$.
In an exhaustive category, transfinite unions preserve monomorphisms. That is, given a transformation $(A_\alpha \to B_\alpha)_{\alpha\lt\kappa}$ of transfinite sequences of monomorphisms which is pointwise monic, also the induced map on colimits $A_\kappa\to B_\kappa$ is monic.
Last revised on February 23, 2013 at 07:46:53. See the history of this page for a list of all contributions to it.