objects such that commutes with certain colimits
Let be an extensive category, then:
By definition, preserves binary coproducts if the canonically defined morphism is always a bijection.
By this definition, the initial object of is not in general connected (except for degenerate cases of ); it is too simple to be simple. This matches the notion that the empty space should not be considered connected, discussed at connected space.
Let be an infinitary extensive category, then
The “only if” is clear, so we just prove the “if”.
We first show that preserves the initial object . Indeed, if preserves the binary product , then the canonical map
is a bijection of sets, where the restriction to is also a bijection of sets . This forces the set to be empty.
Now let be a set of objects of . We are required to show that each map
factors through a unique inclusion . By infinite extensivity, each pullback exists and the canonical map is an isomorphism. We will treat it as the identity.
Now all we need is to prove the following.
Indeed, for each , the identity map factors through one of the two summands in
because preserves binary coproducts. In others words, either or (and the other is ). We cannot have for every , for then would be , contradicting the fact that . So for at least one . And no more than one , since we have whenever .
From the proof above, we extract a result useful in its own right, giving an alternative definition of connected object.
If is connected and is a coproduct decomposition, then the arrow , factors through one of the coproduct inclusions of . If it factors through say , then the subobject is all of , and is forced to be initial by disjointness of coproducts.
Turning now to the if direction, suppose is a map, and put , . By extensivity, we have a coproduct decomposition . One of , is initial, say , and then we have , meaning that factors through , and uniquely so since is monic. Hence belongs to (exactly) one of the two subsets , .
Because coproducts in commute with limits of connected diagrams.
If is connected and is an epimorphism, then is connected.
Certainly is not initial, because initial objects in extensive categories are strict. Suppose (see theorem 2 above), so that we have an epimorphism . By connectedness of , this epi factors through one of the summands, say . But then the inclusion is epi. This forces to be initial.
It need not be the case that products of connected objects are connected. For example, in the topos -Set, the product decomposes as a countable coproduct of copies of . (For more on this topic, see also cohesive topos.)
We do have the following partial result, generalizing the case of .
Suppose is a cocomplete -extensive category with finite products. Assume that the terminal object is a separator, and that Cartesian product functors preserve epimorphisms. Then a product of finitely many connected objects is itself connected.
In the first place, is connected. For suppose where is not initial. The two coproduct inclusions are distinct by disjointness of sums. Since is a separator, there must be a map separating these inclusions. We then conclude , and then by disjointness of sums.
Now let and be connected. The two inclusions are distinct, so there exists a point separating them. For each , the +-shaped object is connected (we define to be a sum of connected objects , amalgamated over the connected object , i.e., to be a connected colimit of connected objects). We have a map
where the union is again a sum of connected objects amalgamated over , so this union is connected. The map is epic, because the evident map
is epic (by the assumptions that is a separator and preserves epis), and this map factors through . It follows that the codomain of is also connected.
The same method of proof shows that for an arbitrary family of connected spaces , the connected component of a point in the product space contains at least all those points which differ from in at most finitely many coordinates. However, the set of such points is dense in , so must also be connected.