I am new to this and would like to understand $0 \overset{a0}{\to} B \overset{a1}{\to} A \overset{a2}{\to} A/B \overset{a3}{\to} 0, $ where $B \subset A$ and they are both Abelian groups. Also maybe it is helpful to know that every subgroup of an Abelian group is normal? This is in the context of a cochain (we were taught starting this direction).
So here is what I understand so far:
If it is a short exact sequence (s.e.s.) (is this given or is it just iff $a_0$ is injective AND $a_3$ is surjective?), then, with the identity element of an additive (for instance) Abelian group being $0$, it is such that,
$\begin{eqnarray} ker\; a_0 \leq_{sg} 0 \;\Rightarrow ker\; a_0=0 \\ ---------------- {} \\ im\; a_1 = ker\; a_2 \text{ and } ker\; a_2 = B \\ H^2(A) = \frac{ker \; a_2}{im\; a_1} = \frac{B}{B} = 0 \\ ----------------- {} \\ a_2: A \to A/B \hspace{1cm} a_2(a_i) = a_i\,B \\ ker\; a_2 = \{a\in A : a_2 \circ a = 0\} = \{a \in A : a_2\circ a = B\} = B\\ im\; a_2 \cong A/B \\ ---------------- {} \\ a_3\; \text{ zero map } \Rightarrow a_3(a\,B) = B \hspace{1cm}\text{ as the zero of $A/B$ is $B$ ? } \\ H^3(A/B) = \frac{ker\; a_3}{im \; a_2} = B/(A/B) \end{eqnarray}$
I'm not sure of these results, but I hope it is simple and clear enough to get some feedback. Thanks!
Edit: I forgot to mention that with this construction, $a_1$ induces an isomorphism from $B \to (\underset{s.g.}{{\cdot}}\subset A)$ and $a_2$ induces an isomorphism from $A/im\;B \to A/B$.