Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In topology a cover of a space is a set of subspaces whose union is the space. Obviously a subspace is an inclusion.

A space is an object in the category $Top$.

In category theory, according to nlab, a cover of an object $U$ (of some category) is a family of morphisms $f_i:U_i\rightarrow U$. It doesn't look as though they require that the 'union' of the $U_i$ needs to be $U$. Nor that they should be 'inclusions'. But if we do require it, what is the best way to generalise the condition? Is there standard nomenclature for this?

The inclusion of a subspace $U_i$, is probably best generalised by requiring that each $f_i$ is monic.

For the 'union' of the $U_i$, can we simply say that the maps are jointly epic? That is whenever $g\circ f_i=h\circ f_i$ for all $i$, we have $g=h$. But this simply says that $U$ is a coproduct of the family $f_i$.

Or alternatively, in topology, we can take the disjoint union of the cover and this will be surjective and locally injective - that is a local homeomorphism. And this is also called an etale mapping.

If the category has all pushouts and an initial object $0$, we can construct the disjoint pushout $\sqcup g_i$ by the initial morphisms $g_i:0\rightarrow U_i$, and then by the pushout property we have $\sqcup g_i:\sqcup U_i\rightarrow U$ that is epic. But in the presence of initial objects pushouts & coproducts can construct each other. So the alternative is actually no real alternative.

share|improve this question
2  
Not every jointly epic family is a coproduct diagram ... –  Martin Brandenburg Jan 5 at 23:41
    
@Brandenburg: ok, its the reverse that is true. Is there a nice condition that makes them equivalent? Actually - I don't think there is, its the 'existence' in the coproduct thats the the problem not the 'uniqueness'. –  Mozibur Ullah Jan 6 at 0:00
    
In $\mathsf{Set}$, $\{X_i \to X\}$ is a coproduct diagramm if it is jointly epic, each $X_i \to X$ is monic, and $X_i \times_X X_j = 0$ (initial) for all $i \neq j$. Probably the same holds for large classes of categories (regular categories?). –  Martin Brandenburg Jan 6 at 0:28
    
@MartinBrandenburg You should probably assume extensiveness as well. Then it becomes a question that can be solved by descent theory. –  Zhen Lin Jan 6 at 2:03
add comment

1 Answer 1

The correct approach is not to internalize the notion of "surjective", but rather to axiomatize the characteristic properties of open coverings. This has been done by Grothendieck, who introduced Grothendieck topologies. He defined them for the purpose of étale cohomology.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.