The connecting morphism exists and is unique

Because the top row is short exact, [L1] says that $p$ is epic and $i=\ker (p)$. Form the pullback of $p$ along $k_h$:

Form the pushout of $q_f$ and $i^{\mathrm{\prime }}$:

Step 1.1 gives $j=\ker (\pi )$ and makes $\pi$ epic. By [L4], the sequence $0\to A\stackrel{j}{\to }P\stackrel{\pi }{\to }K\to 0$ is therefore short exact. Applying [L1] to this new short exact sequence shows that $\pi$ is also a cokernel of $j$.

The pullback relation gives $hp{\pi }^{\mathrm{\prime }}=h{k}_h\pi =0,$ so the snake-data square yields $p^{\mathrm{\prime }}g{\pi }^{\mathrm{\prime }}=0.$ Because $i^{\mathrm{\prime }}=\ker (p^{\mathrm{\prime }})$ by [L1], [L5] gives a unique map $a:P\to A^{\mathrm{\prime }}$ with $i^{\mathrm{\prime }}a=g{\pi }^{\mathrm{\prime }}\mathrm{.}$

Since ${\pi }^{\mathrm{\prime }}j=i$ and the left square commutes, we have $i^{\mathrm{\prime }}aj=g{\pi }^{\mathrm{\prime }}j=gi=i^{\mathrm{\prime }}f\mathrm{.}$ The map $i^{\mathrm{\prime }}$ is monic, so $aj=f$. Therefore $q_{f}aj=q_{f}f=0.$ Because $\pi$ is a cokernel of $j$ by step 2.1, [L5] yields a unique morphism $\delta :K\to Q$ with $\delta \pi =q_{f}a\mathrm{.}$

Composing with the pushout coprojection gives $\iota \delta \pi =\iota q_{f}a={\iota }^{\mathrm{\prime }}{i}^{\mathrm{\prime }}a={\iota }^{\mathrm{\prime }}g{\pi }^{\mathrm{\prime }},$ which is the required relation. If ${\delta }_1$ and ${\delta }_2$ both satisfy that relation, then $\iota {\delta }_1\pi =\iota {\delta }_2\pi \mathrm{.}$ Since $\pi$ is epic by step 1.1 and $\iota$ is monic by step 1.2, this forces ${\delta }_1={\delta }_2$.

Hence the connecting morphism exists and is unique.