Title: In an abelian category, show that $f(A) = f(B)$ implies $A \subseteq ((A+B) \cap \text{Ker} f) + B$.
I'm trying to show this result with as little machinery as possible. This includes not using any diagram lemmas (as far as it is possible) or embedding theorems. This means basically only using universal properties or (simple) facts about subobjects, some of which will be outlined below.
Here $f : X \to Y$ is a morphism, $A$ and $B$ are subobjects of $X$ (equivalence classes of monos with codomain $X$).
$f(A)$ is defined to be the image of the inclusion of $A$ into $X$ followed by $f$.
Sums (unions) and intersections of subobjects are defined as joins and meets in the poset of subobjects. To actually construct these, we can set
$$ A + B = \text{Im}(A \oplus B \to X ) $$
where the maps from $A$ and $B$ to $X$ are just the subobject inclusions.
The intersection can be constructed as the pullback of the diagram consisting of the inclusions of $A$ and $B$ in $X$.
Some extra information that might be useful:
For a morphism $f : X \to Y$ and a subobject $B \subseteq Y$, we can define the preimage $f^{-1}(B)$ to be the largest subobject $P$ of $X$ satisfying $f(P) \subseteq B$, i.e. a subobject $P$ such that $A \subseteq P \iff f(A) \subseteq B$. The preimage can be constructed by taking the pullback of the diagram consisting of f and the inclusion of $B$ into $Y$.
Here are some results I already know (they can be proved elementarily):
$f$ mono $\implies f^{-1}(f(A)) = A$
$f$ epi (or even just $B \subseteq \text{Im} f$) $\implies f(f^{-1}(B)) = B$
$f(A_1 + A_2) = f(A_1) + f(A_2)$
$f^{-1}(B_1 \cap B_1) = f^{-1}(B_1) \cap f^{-1}(B_2)$
Some motivation: this (title) result is supposed to facilitate a diagram chasing technique where one chases subobjects. The one thing I need to make things work out nicely is a notion of "subtraction" of subobjects. More precisely, in a situation where $f(A) = f(B)$, I'd like there to be a subobject, call it $A -_f B$, with is supposed to be a "difference with respect to $f$", which satisfies $f(A -_f B) = 0$ and $A \subseteq (A -_f B) + B$ (equality in the second condition is too much to ask for). In module categories one can check that $(A + B) \cap \text{Ker} f$ does the job, and the title question is just checking that the second condition for this choice of "difference" holds in abelian categories.
One more thing: I am aware of "generalized elements" from Mac Lane's book, but as far as I am aware his "difference" of elements doesn't have a corresponding "$(a - b) + b = a$" property.
This is my first Math StackExchange question, so I apologize for any formatting mistakes or things that may be unclear. Thanks in advance!
Let $V$ be the pullback of $f |_A : A \to f(A) = f(B)$ and $f |_B : B \to f(B)$, with projections $a : V \to A$ and $b : V \to B$. Then $a - b$ factors through $(A + B) \cap \ker(f)$, so $a$ factors through $((A+B) \cap \ker(f)) + B$. However, since $a$ is an epimorphism (as a pullback of an epimorphism), that implies the desired result.