= Solution
Let $Z_t$ denote the generation-$t$ type-count <vector>, and let $p_i^{(t)}=\Pr_i(|Z_t|>0)$ from a single type-$i$ ancestor. Every type-$j$ child independently has <probability> $p_j^{(t)}$ of having descendants in generation $t$. For a <Poisson random variable> $N$ with <mean> $\lambda$, $\mathbb E s^N=\exp[\lambda(s-1)]$. Independence across child types therefore gives
$$
p^{(0)}=\mathbf1,\qquad p_i^{(t+1)}=1-\prod_j\exp(-\lambda_{ij}p_j^{(t)})=T_i(p^{(t)}),\qquad T_i(x)=1-e^{-(\Lambda x)_i}.
$$
The events of nonempty successive generations decrease. Each individual has finitely many children almost surely, so the total tree is infinite precisely when no generation is empty. Consequently $p^{(t)}\downarrow p$, the actual <survival probability of a branching process>. Continuity of $T$ yields
$$
\boxed{p_i=1-\exp\left(-\sum_j\lambda_{ij}p_j\right).}
$$
The map $T$ is coordinatewise increasing on nonnegative <vectors>. Every nonnegative <fixed point> $p'$ automatically lies in $[0,1]^l$, since its coordinates equal $1-e^{-z}$ with finite $z\ge0$. Hence $p'\le\mathbf1$ and, by induction, $p'=T^t(p')\le T^t(\mathbf1)=p^{(t)}$. Taking limits proves \b[$p'\le p$ in every coordinate], so $p$ is the greatest nonnegative solution, not merely some <fixed point>. This argument applies without irreducibility assumptions to the <multitype Poisson branching process>.
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