# (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?

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 $$D$$.