A statement on exact sequences of R-modules. My question is related to the answer of this post:
Exact sequence of $A$-modules.
I am trying to understand this answer, but I am stuck at the following statement.
Given is a sequence $A\stackrel {f}{\longrightarrow } B\stackrel {g}{\longrightarrow } C\longrightarrow 0$ of left R-modules. Given is that $g$ "induces" (what does this mean in this context?) an isomorpism $\mbox{coker } (f) = B/\mbox{im } (f)\cong C$. Prove that $A\stackrel {f}{\longrightarrow } B\stackrel {g}{\longrightarrow } C\longrightarrow 0$ is exact, i.e., $g$ is surjective and $\mbox{im } (f) = \ker (g)$. Can anybody help me to prove this? Because I do not see it.
Is there a dual statement?
 A: If $h:P\to Q$ is a homomorphism of some sort (of groups, modules, rings, etc.), then the statement that $h$ induces a homomorphism $\widetilde{h}:P/S\to Q$ for a given subobject $S$ of $P$ (subgroup, submodule, ideal, etc.) implicitly says that $S\subseteq\ker(h)$, and that $\widetilde{h}$ is defined on cosets by $\widetilde{h}(p+S)=h(p)$ for any $p\in P$.
The key facts are that


*

*$\widetilde{h}$ is injective $\iff$ $S=\ker(h)$, and 

*$\widetilde{h}$ is surjective $\iff$ $h$ is surjective
so if we're told that $h$ induces an isomorphism $P/S\cong Q$, then we know that $S=\ker(h)$ and that $h$ is surjective (or in other words that $Q=\mathrm{im}(h)$).
Thus, if you're given a sequence of maps of left $R$-modules $$A\xrightarrow{\;\;f\;\;}B\xrightarrow{\;\;g\;\;}C\longrightarrow 0$$
and told that $g$ induces an isomorphism $B/\mathrm{im}(f)\cong C$, then by the two facts above, you can conclude that $\mathrm{im}(f)=\ker(g)$ and that $C=\mathrm{im}(g)$ (i.e., that $g$ is surjective). 
A: "g "induces"" means that the map $$\begin{matrix}B/Im(f)&\longrightarrow&C\\\overline{x}&\longmapsto&g(x)\end{matrix}$$ is an isomorphism.
Now, given $y\in Im(f)$ then $\overline{y}=\overline{0}$, so $\overline{y}\longmapsto 0=g(y)$. Therefore $Im(f)\subseteq Ker(g)$. For other hand, if $x\in Ker(g)$, then $g(x)=0\in C$. Thus, $\overline{x}$ belongs to kernel of the isomorphism. Hence $x\in Im(f)$ and so $Ker(g)\subseteq Im(f)$.
Since $Im(f)=Ker(g)$, then the sequence is exact in $B$ and $B/Ker(g)=B/Im(f)$ is isomorphic to $C$ and $Im(g)$. Therefore $C=Im(g)$. It follows that the sequence is exact.
