Given a pointed category $\mathsf C$, let $\mathsf{Pt}(\mathsf C)$ be the category whose objects are pairs $(f,s)$ with $s$ a section of $f$, and arrows are serially commutative squares between such pairs. Let $\mathrm N_0:\mathsf{Pt}(\mathsf C)\to \mathsf C\times \mathsf C$ take $(f,s)$ to $(\operatorname{cod}f,\operatorname{Ker}f)$.
I am struggling with the following proposition from Bourn's Moore Normalization and Dold-Kan for Semi-Abelian Categories.
Proposition 1.1. $\mathrm N_0:\mathsf{Pt}(\mathsf C)\to \mathsf C\times \mathsf C$ is defined and that $\mathsf C$ admits finite products. Then the functor $\mathrm N_0$ has a left adjoint $\Sigma_0$ iff $\mathsf C$ has coproducts.
Given coproducts, I understand the left adjoint should take $(A,B)$ to $((1,0):A\amalg B\leftrightarrows A:\iota_1)$. For the converse the author references Proposition 4 of his paper Normalization Equivalence, Kernel Equivalence, and Affine Categories. I don't understand how this proposition proves the existence of coproducts from existence of the left adjoint in our setting. I'm trying to mimic its proof by parts, namely to prove the left square below is a pushout by producing a unique factorization for every commutative diagram as on the left below. By moving from $C$ to $A\times C$ we can obtain another point over $A$ as in the diagram on the right, whence the adjunction provides the broken arrow.
My problem is that I don't know how to prove the resulting $\pi_2\circ (1_A,b)^\sharp:\Sigma_0(A,B)\to C$ actually factors $b$ as well, i.e how to prove $$\pi_2\circ (1_A,b)^\sharp\circ \ker \sigma_0\circ \eta_{(A,B)} =b.$$ Writing $(1_A,b)^\sharp$ concretely as the counit composed on $\Sigma_0(1_A,b)$ does not seem to help. What to do?
The second component of the right adjoint $N_0$ amounts to taking the restriction to the kernels; and in particular, the kernel of $\pi_A:A\times C\to A$ is simply $(0,1):C\to A\times C$. Let us denote $\kappa : Ker(\sigma_0)\to C$ the restriction of $(1_A,b)^{\sharp}$ through the kernels, i.e. the only map such that $$(0,1)\circ \kappa=(1_A,b)^{\sharp}\circ \ker(\sigma_0).$$ Then $\kappa$ is the second component of the image of $(1_A,b)^\sharp$ by the right adjoint. Composing both sides by $\pi_C$ and $\eta_B$ gives you $$\kappa\circ \eta_B=\pi_C\circ (0,1)\circ \kappa\circ \eta_B=\pi_C\circ (1_A,b)^{\sharp}\circ \ker(\sigma_0)\circ \eta_B.$$
Your map $(1_A,b)^{\sharp}$ is defined by the adjunction, so that applying the right adjoint and composing with $\eta$ sould give you back $(1_A,b)$. In particular, the second component of this is exactly the left-hand side in the equation above, which must then be equal to $b$, and then the right-hand side must be equal to $b$.