As $\pi$ is an epimorphism and $\iota$ is a monomorphism by  Lemma \ref{lemma-cartesian-cocartesian}, uniqueness of $\delta$ is clear. Let $p=y\times_z\Ker(\gamma)$ and $q=\Coker(\alpha)\amalg_uv$.  Let $h:\Ker(\beta)\to y$, $i:\Ker(\gamma)\to z$ and  $j:\Ker(\pi)\rightarrow p$ be the canonical injections.  Let $p:u\to\Coker(\alpha)$ be the canonical projection.  Keeping in mind Lemma \ref{lemma-cartesian-cocartesian} we get a commutative  diagram with exact rows  $$ \xymatrix{ 0 \ar[r] & \Ker(\pi) \ar[r]^j & p \ar[r]^{\pi} \ar[d]_{\pi'} & \Ker(\gamma) \ar[d]_i \ar[r] & 0 \\ & x \ar[r]^f \ar[d]_\alpha & y \ar[r]^g \ar[d]_\beta & z \ar[d]_\gamma \ar[r] & 0 \\ 0 \ar[r] & u \ar[r]^k \ar[d]_p & v \ar[r]^l \ar[d]_{\iota'} & w & \\ 0 \ar[r] & \Coker(\alpha) \ar[r]^\iota & q & & } $$ As $l \circ \beta \circ \pi' = \gamma \circ i \circ \pi = 0$ and as the third  row of the diagram above is exact, there is an $a:p\to u$  with $k \circ a = \beta \circ \pi'$. As the upper right quadrangle of the  diagram above is cartesian, Lemma \ref{lemma-cartesian-kernel} yields an  epimorphism $b : x \to \Ker(\pi)$ with $\pi' \circ j \circ b = f$.  It follows  $k \circ a \circ j \circ b = \beta \circ \pi' \circ j \circ b = \beta \circ f = k \circ \alpha$. As $k$ is a monomorphism this implies $a \circ j \circ b = \alpha$. It follows  $p \circ a \circ j \circ b = p \circ \alpha = 0$. As $b$ is an epimorphism this  implies $p\circ a\circ j=0$. Therefore, as the top row of the diagram  above is exact, there exists $\delta : \Ker(\gamma) \to \Coker(\alpha)$ with $\delta \circ \pi = p \circ a$. It follows  $\iota \circ \delta \circ \pi = \iota \circ p \circ a = \iota' \circ k \circ a = \iota' \circ \beta \circ \pi'$ as desired. \medskip\noindent As the upper right quadrangle in the diagram above is cartesian there  is a $c : \Ker(\beta) \to p$ with $\pi' \circ c = h$ and $\pi \circ c = g'$.  It follows  $\iota \circ \delta \circ g' = \iota \circ \delta \circ \pi \circ c = \iota' \circ \beta \circ \pi' \circ c = \iota' \circ \beta \circ h = 0$.  As $\iota$ is a monomorphism this implies $\delta \circ g' = 0$. \medskip\noindent Next, let $d : r \to \Ker(\gamma)$ with $\delta \circ d = 0$. Applying Lemma \ref{lemma-check-exactness-fibre-product} to the exact sequence  $p \overset{\pi}\to \Ker(\gamma) \to 0$ and $d$ yields an object $s$,  an epimorphism $m : s \to r$ and a morphism $n : s \to p$ with  $\pi \circ n = d \circ m$. As $p \circ a \circ n = \delta \circ d \circ m = 0$,  applying Lemma \ref{lemma-check-exactness-fibre-product} to the exact sequence  $x \overset{\alpha}\to u \overset{p}\to \Coker(\alpha)$ and  $a \circ n$ yields an object $t$, an epimorphism $\varepsilon : t \to s$ and  a morphism $\zeta : t \to x$ with $a \circ n \circ \varepsilon = \alpha \circ \zeta$.  It holds  $\beta \circ \pi' \circ n \circ \varepsilon = k \circ \alpha \circ \zeta = \beta \circ f \circ \zeta$.  Let $\eta = \pi' \circ n \circ \varepsilon - f \circ \zeta : t \to y$. Then,  $\beta \circ \eta = 0$. It follows that there is a  $\vartheta : t \to \Ker(\beta)$ with $\eta = h \circ \vartheta$. It holds  $i \circ g' \circ \vartheta = g \circ h \circ \vartheta = g \circ \pi' \circ n \circ \varepsilon - g \circ f \circ \zeta = i \circ \pi \circ n \circ \varepsilon = i \circ d \circ m \circ \varepsilon$.  As $i$ is a monomorphism we get $g' \circ \vartheta = d \circ m \circ \varepsilon$.  Thus, as $m \circ \varepsilon$ is an epimorphism, Lemma \ref{lemma-check-exactness-fibre-product} implies that  $\Ker(\beta) \overset{g'}\to \Ker(\gamma) \overset{\delta}\to \Coker(\alpha)$  is exact. Then, the claim follows by Lemma \ref{lemma-exact-kernel-sequence} and duality.