Compatibility of homomorphisms and quotient maps of abelian groups Suppose $A$ and $C$ are abelian groups with subgroups $A'$ and $C'$ respectively. Let $f:A\to C$ be a group homomorphism. I was wondering if the following statements are equivalent:


*

*There exists a homomorphism $g:A'\to C'$ such that $fi=jg.$

*There exists a homomorphism $h:A/A'\to C/C'$ such that $hp=qf.$


$$\begin{array}
AA' & \stackrel{i}{\longrightarrow} & A & \stackrel{p}{\longrightarrow} & A/A'\\
\downarrow{g} & & \downarrow{f} & & \downarrow{h} \\
C' & \stackrel{j}{\longrightarrow} & C  & \stackrel{q}{\longrightarrow} & C/C'
\end{array}
$$
where $i$ and $j$ are inclusions, and $p$ and $q$ are projections. 
It seems to me that (1) means the restriction of $f$ to $A'$ is g, so to prove (1) implies (2), I define $h(a+A')=f(a)+g(A')$. But $g(A')\neq C'$ in general, and I'm confused about if I made the right definition.
 A: Yes, these statements are equivalent, and you're approach is right. Note only that in order to describe an element of $C/C^{\prime}$, when the latter is realized by $C^{\prime}$-cosets in $C$, you have to put $h(a + A^{\prime}) := f(a) + C^{\prime}$. However, more conceptually you should try to prove the equivalence using universal properties of kernel and cokernel only - this has the advantage of directly generalizing to arbitrary abelian categories.
A: For (2) $\Rightarrow$ (1):
Take $a'\in A'$, we want to define an element in $C'$.  The only map available to use is $i:A'\to A$ which gets you to $A$.  As I mentioned in the comment, one way to get something in $C'$ is to get $[0]=0+C'\in C/C'$.  So we push our element $i(a')\in A$ to element in $C/C'$ there is only one way to do it: so you have $qfi(a')=hpi(a)\in C/C'$.
Note that $pi=0$, so $hpi(a)=q(fi(a))=[0]$.  Hence $fi(a)\in \ker(q)=\text{im}(j)$.  That is, there is some $c'\in C'$ such that $j(c')=fi(a)$.  
If there is another $c''\in C'$ with $j(c'')=fi(a)$, then $j(c'')=j(c')$, but $j$ injective, so $c''=c'$.  Hence, the assignment $h:a'\mapsto c'$ is well-defined. In particular, you get (1).
(1) $\Rightarrow$ (2) is the "dual" argument; I will leave it to you to attempt again.
When you finished the exercise, it is perhaps good to have a look at "pullback and pushforward".  You can find them in most homological algebra textbook.
