Expressing "a subobject is a union of other subobjects" in a topos

Let $(\mathsf C,J)$ be a site. Let $U\rightarrowtail X$. Suppose there's a family of subobjects $(U_i\rightarrowtail X)$ which factors through $U\rightarrowtail X$ such that the factorized family $$(U_i\rightarrowtail U)$$ is a $J$-covering family.

If the site is superextensive, is it true $\coprod U_i\to U$ is an isomorphism?

The map $\coprod U_i \rightarrow U$ will in general be a cover, that is an epimorphism in the topos. But it has no reasons to be a monomorphism, basically the $U_i$ can overlap.

A trivial example of this you can have only two $U_i$ with $U_1=U_2=U$.

$U_1,U_2 \rightarrow U$ is obviously a cover, but $U_1 \coprod U_2 \rightarrow U$ is not going to be an isomorphism unless $U$ is the initial object.

What is true (without any assumption on the site), is that in the topos, the object represented by $U$ can be written as a colimit of the diagram with all the $U_i$ and all the $U_i \times_{U} U_j$ with maps the projections from $U_i \times_{U} U_j$ to $U_i$ and $U_j$. This corresponds to gluing the $U_i$ together but identifying their intersections/overlap ($U_i \times_U U_j$ is basically the intersection of $U_i$ and $U_j$)

The proof of this is almost trivial: if you wrote down the universal property of this colimit for morphism into an object $X$ of the topos you get exactly the sheaf condition for $X$ and the covering of $U$ by the $U_i$.

If the site is subcanonical and the those fiber products exists in the site then this colimit also holds in the site.

If the site is subcanonical and the fiber product are not necessarily present in the site you still get a colimit of this kind by replacing $U_i \times_U U_j$ by some class of object that have map to the $U_i \times_U U_j$ exactly as when forulating the sheaf condition on a site where there is no fiber product.

If the site is not subcanonical you can basically never deduce anythings on its underlying category from properties of the topos, so there is nothing you can do here.

• Ah, silly me. Accidentally looked at the coproduct as a union. Thanks!! Oct 19 '16 at 14:42