Is the connecting homomorphism
$ d^* : H^k(\mathcal C) \to H^{k+1}(\mathcal A) $ well defined?
$0 \to \mathcal A \to \mathcal B \to \mathcal C \to 0$ is short exact.
$\mathcal A$, $ \mathcal C$ are cochain complexes, while $H^k(\mathcal X) = \frac{\ker(d^\mathcal X_k)}{\text{image}(d^\mathcal X_{k-1})}$
I thought that the way to go is to show that, given $c, c' \in [c]\in H^k(\mathcal C), c, c' \in C_k$ then $d*[c-c'] = 0$.
Since $j$ is onto, $c-c' = j(b)-j(b') = j(b-b')$, with $b, b' \in B_k$. Then $0 = d(c-c') = j(db-db')$. So $db-db'\in \ker(j) = \text{image}(i)$, which is one-to-one. There is a unique $a^* \in A^{k+1}$ such that $i(a^*) = db-db'$. If only I could show that $(b-b') = d(b''), b'' \in B_{k-1}$ I could conclude that $di(a*) = id(a*) = 0 \iff da^* = 0 \iff a^* \in \text{image}(d^{\mathcal A}_{k})$.
Is that correct?