In a commutative diagram, lifting the image via a surjective map I have $R$-modules $A_1,A_2,B_1,B_2,C$ such that the following diagram commutes,
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{lll}
A_2 & \ra{\partial} & A_1 \\
\da{f} & & \da{g} \\
B_2 & \ra{\partial} & B_1 \\
\end{array}\;,
$$
and such that the following is an exact sequence,
$$
 0 \rightarrow C \xrightarrow{i} A_1 \xrightarrow{g} B_1 \rightarrow 0~.
$$
The maps $\partial$ are induced by the usual boundary maps, and $i$ is the inclusion map.
My objective is to find,
$$H = B_1 / \partial(B_2)~.$$
Then, using the exact sequence, I can say that,
$$
 B_1 \sim A_1 / C~.
$$
Could you please tell me what are the sufficient conditions that maps $f$ and $g$ need to satisfy such that I can lift the image $\partial(B_2)$ to $\partial(A_2)$ in $A_1$, and write the following?
$$
 H = B_1 / \partial(B_2) \sim A_1 / (C +  \partial(A_2))~.
$$
Thank you. This is not for homework. I saw this in a paper where $f$ and $g$ were both surjective, but I don't understand the reasoning behind the step. If $f$ and $g$ are both surjective and injective, then it would obviously hold; my question is if a weaker condition would be sufficient. In particular, I would be very interested in when $g$ can certainly be no more than surjective, i.e., it must have a non-trivial kernel.
 A: This is basically a result about cokernels.  And yes, you need both $f$ and $g$ to be surjective for this to work.  
Your assumptions give you two exact sequences, one $0 \to \mathrm{ker}f \to A_2 \to B_2 \to 0$ and another obvious one for $g$.  Draw the former on top of the latter and note that the $\partial$ maps provide well-defined (vertical) homomorphisms between the kernels, the $A$s, and $B$s.  You should check this, especially at the kernels.  
Now take the vertical cokernels of the maps $\partial: A_2 \to A_1$ and $\partial: B_2 \to B_1$.  The second cokernel is the one you wish to compute.  There is an obvious map induced by $g$,
$$
g_*:  A_1/\partial(A_2) \to  B_1/\partial(B_2)
$$
via $g_*(a + \partial(A_2))= g(a) + \partial(B_2)$.  This is well-defined, and is clearly onto since $g$ was.  By the first iso theorem, $A_1/\partial(A_2)$ mod the kernel of $g_*$ is $B_1/\partial(B_2)$, which is what you want.
So, what's the kernel?  If $a \in A_1$ has $g(a) \in \partial(B_2)$ then $g(a) =\partial(b)$ for some $b \in B_2$.  Since $f$ is onto, there is $a' \in A_2$ with $f(a')=b$.
Claim:  $a-\partial{a'} \in \mathrm{ker}g$.
Well, yes, because $g(a-\partial(a')) = g(a) - g\partial(a') = \partial(b) - g\partial(a') = \partial(b)-\partial f(a')=\partial(b)-\partial(b)=0$.
Rephrased, $a= c + \partial(a')$ where $c$ is an element of $\mathrm{ker}g$.  The converse also holds.
What you just learned:  the kernel of $g_*$ consists of all cosets of the form $c + \partial(A_2)$ where $c \in \mathrm{ker}g$.  In short,
$$
\mathrm{ker}(g_*) = \mathrm{ker}g + \partial(A_2).
$$
I guess in your notation, this is $C + \partial(A_2)$.  Now you're done, by either the second or third isomorphism theorem (I never remember which is which).
