The induced morphism $B/A\to C/A$ is monic/epi if the morphism $B \to C$ is monic/epi in abelian categories Let $f: A \to B$ and $g: A \to C$ be two monomorphisms in an abelian category. By definition, $B/A:= \operatorname{coker}(f)$, $C/A:= \operatorname{coker}(g)$.
If $h:B\to C$ is another morphism such that $h\circ f=g$, then by definition of cokernel, there exists a unique map $h':B/A \to C/A$ making the diagram commute.
My question is, does $h$ monic/epi imply $h'$ monic/epi? How to prove it without invoking Mitchell's embedding theorem(which requires the category to be small)? 
In the category of $R$-modules, this is true(so by the embedding theorem this should be true for small abelian categories). But I want to know how to use an element-free argument to prove it(for categories that are big or small).
 A: You don't really need the snake lemma for this.
Your situation can be described by the following diagram
$$\require{AMScd}
\begin{CD}0 @>>> A @>{f}>> B @>{\operatorname{coker}(f)}>> \frac{B}{A} @>>> 0 \\
 &  @V_{id_A}VV & @VV{h}V *& @VV{h'}V\\
0 @>>> A @>>{g}>C @>>{\operatorname{coker}(g)}> \frac{C}{A} @>>> 0 
\end{CD}$$
where both rows are short exact sequences. Then because the kernels are isomorphic, the square $*$ is a pullback (see for example this question), and pullbacks in abelian categories reflect monomorphisms, so if $h$ is a mono, then so is $h'$.
The case where $h$ is an epimorphism is easier : since cokernels are epimorphisms as well, $h'\circ \operatorname{coker}(f)=\operatorname{coker}(g)\circ h$ is an epimorphism, and thus $h'$ is an epimorphism.
A: I will illustrate a method of generalizing diagram chasing arguments to arbitrary abelian categories.  So, let us start with the following argument in the case of abelian groups: suppose $h$ is monic, and we want to show $h'$ is monic.  It suffices to show $\ker(h') = 0$.  Thus, suppose we have $\bar b \in B / A$ such that $h'(\bar b) = 0$.  Then $\bar b$ has some preimage $b \in B$.  Now, we know that $h(b) \in \mathrm{im}(g)$ by the assumption that $h'(\bar b) = 0$.  Thus, choose $a \in A$ such that $g(a) = h(b)$ (which happens to be unique).  Then $h(f(a)) = g(a) = h(b)$, so $f(a) = b$ since $h$ is monic.  This implies that $\bar b = \overline{f(a)} = 0$.
To generalize this to abelian categories, suppose you have a test object $U$ and an element $\bar b \in (B/A)(U) := \mathrm{Hom}(U, B/A)$ such that $h'(\bar b) := h' \circ \bar b = 0$ in $(C/A)(U)$.  Then there is some epimorphism $\phi : V \to U$ and a "section" $b \in B(V)$ such that $\phi^* \bar b := \bar b \circ \phi \in (B/A)(V)$ is equal to the projection of $b$.  (In particular, $V := U \times_{B/A} B$ works, with $\phi$ and $b$ being the projections to $U$ and $B$ respectively.)  By replacing $U$ with $V$ and $\bar b$ with $\phi^* \bar b$, we may assume $V = U$ and $\bar b$ is equal to the projection of $b$ (since the pullback by an epimorphism is injective).  Now, since the projection of $h(b)$ to $(C/A)(U)$ is zero, this implies that $h(b)$ factors uniquely through $g$; let $a \in A(U)$ be the factorization, so that $g(a) = h(b)$.  Then $h(f(a)) = g(a) = h(b)$, so since $h$ is monic, this implies $f(a) = b$.  But this implies $\bar b = 0$.  Therefore, we have shown that the zero map $0 \to B$ satisfies the universal property of $\ker(h')$, so $\ker(h') = 0$.
(So, the idea is: whenever the diagram-chasing argument makes use of the surjectivity of some map, translate that into replacing the test object $U$ by some other test object $U'$ with an epimorphism $U' \to U$ such that the section in question lifts, and pulling back all "sections" $X(U)$ to $X(U')$ which doesn't change equality of sections.  And of course, if the diagram-chasing argument makes use of the injectivity of some map, that translates into directly using the fact that the corresponding map is monic.  I find it helpful to think of the case where you are working on a category of sheaves of abelian groups, and then repeatedly refining a cover of the original open set $U$ over which sections are defined whenever we need to use the surjectivity of a sheaf morphism.)
(And if the problem were to show some map $\alpha : X \to Y$ is epic, then the translation of the diagram-chasing argument would show: for each test object $U$ and $y \in Y(U)$, there is an epimorphism $V \to U$ and "preimage" $x \in X(V)$ such that $\alpha(x)$ is the pullback of $y$.  Then, applying this in the special case $y = \mathrm{id}_Y \in Y(Y)$, we would get that $\alpha \circ x$ is equal to the epimorphism $V \to Y$, which implies $\alpha$ is an epimorphism.)
