(Sup) property for an abelian category The $(Sup)$ property for abelian category states: For any ascending chain $\Omega$ of sub-objects of an object $M,$ the supremum exists; and for any sub-object $L$ of $M,$ we have isomorphism $$Sup\{X\cap L: X\in \Omega\}=(Sup\Omega)\cap L$$
Let $\mathcal{A}$ be an abelian category with property $(Sup)$ and let $I$ be a directed family of sub-objects of an object $M.$ Then does supremum exists for $I$? 
Moreover, is the morphism $Sup\{X\cap L: X\in I\}\rightarrow(Sup\;I)\cap L,$ an isomorphism for any sub-object $L$ of $M$?  
What about any family of sub-objects of $M$?
I can prove that in case $\mathcal{A}$ is cocomplete, then $(Sup\;I)$ exists for any family, but does the isomorphism holds?
Thanks in advance! 
 A: Note that everything here takes place inside the poset of subobjects of $M$ (which is a lattice since $\mathcal{A}$ is abelian).  We can thus state more generally:

Theorem: Let $P$ be a lattice such that every chain in $P$ has a join.  Then $P$ is complete: every subset of $P$ has a join.  Moreover, if $x\in P$ and $x\wedge -$ preserves joins of chains, then $x\wedge -$ preserves joins of arbitrary directed sets.

Proof: Let $S\subseteq P$.  Fix a well-ordering $\prec$ of $S$, which we extend to a well-ordering of $S\cup\{\infty\}$ by saying $a\prec \infty$ for all $a\in S$.  For each $a\in S\cup\{\infty\}$, let $S(a)=\{b\in S:b\prec a\}$.  We prove by induction on $a$ that the join $\bigvee S(a)$ exists for all $a\in S\cup\{\infty\}$.  Indeed, suppose $\bigvee S(b)$ exists for all $b\prec a$.  If $a$ is a limit, note that $\{\bigvee S(b):b\prec a\}$ is a chain and so its join exists, and that join coincides with $\bigvee S(a)$.  If $a$ is the successor of some element $b$, then $S(a)=S(b)\cup\{b\}$, so $\bigvee S(a)=b\vee\bigvee S(b)$.
In particular, in the case $a=\infty$, we conclude that $\bigvee S(\infty)=\bigvee S$ exists.  Thus $P$ is complete.
Now suppose $S\subseteq P$ is a directed set and $x\in P$ is such that $x\wedge-$  preserves joins of chains.  We prove by (transfinite) induction on $|S|$ that $x\wedge-$ preserves the join of $S$; that is, $x\wedge\bigvee S=\bigvee\{x\wedge a:a\in S\}$.
So, suppose that $x$ preserves the join of $T$ for any set $T\subseteq P$ of cardinality smaller than $|S|$.  If $S$ is finite, then by directedness it has a greatest element, and so it is obvious that $x\wedge-$ preserves the join of $S$.  We may thus assume $S$ is infinite.  But then we can pick a well-ordering $\prec$ of $S$ whose order-type is a cardinal, so that $|S(a)|<|S|$ for all $a\in S$ and $S=\bigcup_{a\in S} S(a)$.  We then see that $\bigvee S=\bigvee \{\bigvee S(a):a\in S\}$.  Since $\{\bigvee S(a):a\in S\}$ is a chain and $x\wedge-$ preserves the join of $S(a)$ for each $a\in S$, we conclude that $x\wedge-$ preseves the join of $S$.

However, it is not typically true that $x\wedge -$ will preserve arbitrary joins, even in the case of a nice abelian category.  For instance, if $\mathcal{A}=Ab$, let $M=\mathbb{Z}\oplus\mathbb{Z}$, let $A=\mathbb{Z}\oplus 0$, let $B=0\oplus\mathbb{Z}$, and let $D$ be the diagonal $\{(n,n):n\in\mathbb{Z}\}$.  Then $D\cap A=D\cap B=0$, but the join of $A$ and $B$ as subobjects of $M$ is all of $M$, whose intersection with $D$ is $D$.
