When is the lattice of subobjects a complete lattice? In any topos $\mathcal{E}$, any object $d$ has a lattice of subobjects $(\operatorname{Sub}(d), \subset)$.
One can give a good description of the object $a\cap b$ for $a,b$ subobjects of $d$, and similarly $a\cup b$. 
This description uses products, so it seems natural, if we want $\operatorname{Sub}(d)$ to be a complete lattice to assume that the topos $\mathcal{E}$ has arbitrary products (although I don't know if it's a necessary condition).
Therefore my questions are : 
Is "$\mathcal{E}$ is closed under arbitrary (small) products" a sufficient condition for "for all $\mathcal{E}$-object $d$, $\operatorname{Sub}(d)$ is a complete lattice" ?
Are there known and interesting necessary and sufficient conditions so that they are complete lattices ?
 A: To talk about intersections of subobjects in general only needs pullbacks of monomorphisms. This doesn't require products to exist. Unions are more complicated in general. In a topos they are pushouts of the projections of an intersection. Of course, a topos has finite products and coproducts. My point is if you want to generalize appropriately, you need to focus on the precise structure needed.
As Kevin Carlson points out, adding all products gives you all limits, and thus certainly arbitrarily wide pullbacks of monomorphisms, but it gives you many other things beyond that. Categories with the structure needed to allow for arbitrary intersections of subobjects are sometimes referred to as being mono-complete, which is indeed exactly that they have wide pullbacks of monomorphisms.  FinSet is mono-complete in this sense getting around the issue Kevin mentioned.
In a topos, or any category with a subobject classifier, we can internalize this into an operation $p=\bigwedge_{i\in I} p_i : A \to \Omega$ for an arbitrary family $\{p_i : A \to \Omega \mid i \in I\}$ such that $\land \circ \langle p,p_i \rangle = p$ for each $i \in I$ and, for all $q : A \to \Omega$, if $\forall i \in I. \land \circ \langle q,p_i \rangle = q$ then $\land \circ \langle q,p \rangle = q$.
I don't think the extra structure of a topos provides any short-cut to having arbitrary intersections. When we talk about "arbitrary" pullbacks/products/limits, we mean set-indexed versions, and the structure of an (elementary) topos gives no tools for talking about this external notion of set or set-indexed families. 
A: Yes, the existence of all products suffices to construct all intersections of subobjects. This is simply because a topos is finitely complete, so the existence of products implies the existence of all limits. Since toposes are finitely cocomplete and a lattice has all intersections whenever it has all unions, we can make the same comment regarding coproducts.
This is certainly not a necessary condition, as the topos of finite sets certainly has complete subobjects lattices without admitting any nontrivial infinite products or coproducts. But this admits a logical embedding into the topos of sets, which does have all products...Perhaps there are better examples that don't come this way, I'm not sure.
