Let $N \rightarrowtail G \twoheadrightarrow H$ be a sequence of group homomorphisms, with maps $\alpha: N \rightarrowtail G$ a monomorphism and $\beta: G \twoheadrightarrow H$ an epimorphism. I am wondering why the sequence is exact iff the following diagram with injective $\alpha$ and surjective $\beta$ is bicartesian, i.e. $N$ in the following is the limit and $H$ is the colimit with $\alpha$ injective and $\beta$ surjective:
\begin{array}{ccc} N & \xrightarrow{\alpha} & G \\ \text{proj}\downarrow & & \downarrow\beta \\ 1 & \hookrightarrow & H \end{array}
I certainly get that the above diagram commutes implies that $\beta \circ \alpha = 1$, i.e. $\text{Im } \alpha \leq \ker \beta$. However, I am not sure how the fact that the square is bicartesian iff $\text{Im } \alpha = \ker \beta$.
If $\iota$ is the inclusion $\ker \beta \hookrightarrow G$, the diagram $$ \require{AMScd} \begin{CD} \ker \beta @>\iota>> G \\ @VVV @VV{\beta}V \\ 1 @>>> H \end{CD} $$ commutes. Since $N$ is the limit, there is a map $\kappa \colon \ker \beta \to N$ such that $$ \alpha \circ \kappa = \iota. $$ Hence: $$ \ker \beta = \iota(\ker \beta) = \alpha(\kappa(\ker \beta)) \leq \operatorname{im} \alpha. $$