Suppose $$ A^\bullet \colon \; \cdots \to A^{i-1} \xrightarrow{f^{i-1}} A^{i} \xrightarrow{f^{i}} A^{i+1} \to \cdots $$ is a complex in an abelian category. I would like to construct an exact sequence $$0 \to H^{i}(A^\bullet) \xrightarrow{t} coker(f^{i-1}) \xrightarrow{p} im(f^{i}) \to 0 $$ To get $p$, just note that the composition of $f^{i-1}$ with the unique morphism $\tilde{f^{i}}\colon A^{i} \to im(f^{i})$ making the triangle involving $f^{i}$ and the kernel of the cokernel of $f^{i}$ commute is zero and thus factors uniquely through the cokernel of $f^{i-1}$. Since $\tilde{f^{i}}$ is an epimorphism, the same is true for $p$.
The morphism $t$ is the unique morphism $H^{i}(A^\bullet) \to coker(f^{i-1})$ whose precomposition with the cokernel of the canonical map $im(f^{i-1})\to ker(f^{i})$ is equal to the composition of the cokernel of $f^{i-1}$ with the kernel of $f^{i}$.
Also, $p\circ t = 0$ because it is zero after precomposing with the cokernel $ker(f^{i})\to H^{i}(A^\bullet)$ of the canonical map $im(f^{i-1})\to ker(f^{i})$.
$\textbf{Question}$ Why is $t$ the kernel of $p$? I don't even see how $t$ is a monomorphism. The problem is that there don't seem to be obvious morphisms going out from $coker(f^{i-1})$ except $p$ and zero morphisms, so I am not able to put it down to something which is a monomorphism (e.g. a kernel of something).
I am grateful for any sort of help. Though, I prefer hints.
As you requested, this is more a hint than an answer, though maybe too much explicit. Your exact sequence is part of the six-term ker-coker exact sequence associated to the obvious map of exact sequences from $0 \to im(f^{i-1}) \to im(f^{i-1}) \to 0 \to 0$ to $0 \to ker(f^i) \to A^i \to im(f^i) \to 0$. So the proof of the ker-coker exact sequence will solve your question.