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.
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).