Projection between quotients by related subobjects For a subobject $A\overset{a}{\rightarrowtail} B$ we define the quotient object $B\twoheadrightarrow B/A$ as the cokernel of any monic representing the subobject.
Suppose we have another subobject $A^\prime\overset{a^\prime}{\rightarrowtail} B$ with $a\leq a^\prime$. I think there should be a natural "projection" between the quotient objects somehow induced by the relation $a\leq a^\prime$: $$B/A\twoheadrightarrow B/A^\prime$$
since intuitively, we "mod out by more" in $B/A^\prime$. This quotient should probably be a factorization of the projection (quotient object) $B\twoheadrightarrow B/A^\prime$ as the two-step projection: $$B\twoheadrightarrow B/A \twoheadrightarrow B/A^\prime$$
However I only have the quotient object $A^\prime /A$... How does the arrow $B/A \twoheadrightarrow B/A^\prime$ come from $A^\prime /A$?

Furthermore, what if I don't even know $a\leq a^\prime$ and I just have some subobject $A\rightarrowtail A^\prime$ which may not factor $a$?
In summary:


*

*Are the things I said up until $B\twoheadrightarrow B/A \twoheadrightarrow B/A^\prime$ correct? How can I prove them?

*In particular, what will the arrow $B/A \twoheadrightarrow B/A^\prime$ be?

*What indeed can I say if I don't know $a\leq a^\prime$ and I just have some subobject $A\rightarrowtail A^\prime$ which may not factor $a$?

 A: Well, as far as I know, the only way to guarantee such an arrow is to induce it by some universal property. In this case, universal property of cokernel. Consider commutative diagram

where $a = a'f$ and $\nu$, $\nu'$ are cokernels (observe that $a = a'f$ forces $f$ to be mono). We have $$\nu'a = \nu' a' f = 0$$ so there is unique map $g\colon B/A\to B/A'$ such that $g\nu = \nu'$ ($g$ is necessarily epi because $g\nu$ is). That should answer 1. and 2., but for 3., I am not sure, but am skeptical that you could induce such a map. Factoring of $a$ through $A'$ was the key property here. Maybe someone else could add some insight in the matter.
EDIT:
Ok, I checked some things and it turns out in Abelian category you can have what you want and it is exactly the third isomorphism theorem! Let me retype what I said in comment. If $\nu'' = \operatorname{coker} f$, we have $\nu a' f = \nu a = 0$ so there is unique map $h\colon A'/A\to B/A$ such that $\nu a' = h\nu''$. Now, consider commutative diagram

where top and middle rows are exact, and columns are exact. Then, by 9-lemma, the bottom row is exact as well, and thus you have $h = \ker g$.
A: (3) is kind of open-ended. I have two things to say.


*

*If $A \overset{a''}{\rightarrowtail} A'$, then $A \overset{a'a''}{\rightarrowtail} B$ factors through $A' \overset{a'}{\rightarrowtail} B$, so you do have the original setup, it's just that the additional subobject $A \overset{a}{\rightarrowtail} B$ is extraneous information.

*If we put the previous point aside, I suppose it's still possible to ask "given $a,a',a''$ as above, does there exist an epimorphism $B/a \twoheadrightarrow B/a'$?". The answer, unsurprisingly, is no. Let $A = \mathbb{Z}$, $A' = \mathbb{Q}$, and $B = \mathbb{Z} \oplus \mathbb{Q}$, with $a,a'$ the obvious inclusions. Then there is an obvious monomorphism $A \to A'$, but the only homomorphism $B/a \to B/a'$ is the zero map, which is not epi.
The dual question (does an arbitrary quotient map between two quotients of $B$ induce a map between their kernels?) has a similarly easy counterexample by taking $B = \mathbb{Z} \oplus \mathbb{Z}/p$.
