Proving two subobjects are equal

Question

I don't think I can really represent this problem properly without quoting precisely what I was given:

Prove that, for any object $A$ in a category $\mathscr{C}$, and any subobjects $M, N \in Sub_\mathscr{C} (A)$, show that $M=N$ if and only if, for every generalized element $x : X \rightarrow A$, $z \in_A M$ if and only if $z \in_A N$.

A footnote: I believe the use of $z$ is a typo and $x$ was intended, but I could be wrong since I don't feel I have the firmest grip on subobjects yet.

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Problem I'm Having #1:

I feel that the double "if and only if's" are tripping me up, I'm not 100% on how to contend with them. My initial instinct is that we're to prove:

• $(\Rightarrow)$ $M=N$ implies, for all subobjects $x : X \rightarrow A$, $x \in_A M$ implies $x \in_A N$, and $x \in_A N$ implies $x \in_A M$.

• $(\Leftarrow)$ For all subobjects $x : X \rightarrow A$, when $x \in_A M$ implies $x \in_A N$, and $x \in_A N$ implies $x \in_A M$, these imply $M =N$.

But I'm very uncertain about this; I think this is the first time I've ever seen double if-and-only-if statements in a theorem.

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Problem I'm Having #2:

If I understand the definition of subobject correctly, then a subobject of $A$ is any monic arrow with codomain $A$, and the local membership relation "$\in_A$" means that $x \in_A M$ if and only if there exists $f : X \rightarrow M$ such that $m \circ f = x$, i.e. the triangle below commutes

I just want to make sure that these definitions are correct because I'm not even 100% where to get started, at least on the latter implication (if I'm somehow right).

Answer 1

Almost everything is correct as you write.
The subobjects are rather equivalence classes of monomorphisms:

We regard monomorphisms $m:M\to A$ and $n:N\to A\ $ as the same subobject of $A$,
$\ \ $ if there is an isomorphism $i:M\to N$ such that $n\circ i=m$.

For the sake of clarity, let $\ \sim\ $ denote this equivalence relation on the set of monomorphisms to $A$.
(By the way, the given notation is very sloppy. If we want to be precise, $M$ is only the domain of a representative monomorphism of the subobject we're talking about; however $M=N$ is used in this context to mean $m\sim n$)

More generally, there's a natural preorder, defining $m\le n$ by the same condition without requiring $i$ to be isomorphism, which corresponds to the usual inclusion of subobjects, and we have $m\sim n\ $ if and only if $\ m\le n$ and $n\le m$.

You can also prove the corresponding statement about generalized elements for inclusions. Then the equality of subobjects will immediately follow.