= Solution
Start the <Galton-Watson process> with one ancestor, $Z_0=1$, and let every individual independently have a number of offspring distributed as $Z$. With independent copies $Z_{t,j}$,
$$
Z_{t+1}=\sum_{j=1}^{Z_t}Z_{t,j},
$$
where an empty sum is zero. Extinction is the event that some generation is empty; zero is then absorbing. Write $q$ for its <probability> and $1-q$ for the survival <probability>.
Set $q_t=P(Z_t=0)$, with $q_0=0$. Conditional on the ancestor's number of children, extinction by generation $t+1$ requires independent extinction by generation $t$ in every child subtree. Thus
$$
q_{t+1}=f_Z(q_t),\qquad f_Z(s)=\mathbb E(s^Z).
$$
The events $\{Z_t=0\}$ increase, so $q_t\uparrow q$. The <probability generating function> is continuous on $[0,1]$, by domination of $s^Z$ by $1$, giving $q=f_Z(q)$. If $r\in[0,1]$ is any <fixed point>, then $q_0=0\leq r$ and monotonicity of $f_Z$ inductively gives $q_t\leq r$. Taking limits proves the <Galton-Watson extinction fixed point> characterization:
$$
\boxed{q=\min\{s\in[0,1]:f_Z(s)=s\}.}
$$
In particular it is the smallest nonnegative <fixed point> wherever the generating series is defined, since it already lies in $[0,1]$.
Let $m=\mathbb EZ=f_Z'(1-)$, allowing $m=\infty$. The <probability generating function> is increasing and convex, with $f_Z(1)=1$. If $m<1$, its tangent inequality gives $f_Z(s)\geq1+m(s-1)>s$ for $s<1$, so $q=1$. If $m=1$ and $P(Z=1)\neq1$, there is positive mass at both zero and some integer at least two; strict convexity gives $f_Z(s)>s$ for $s<1$, again $q=1$. The exceptional case $Z=1$ almost surely has one individual forever, so $q=0$ despite $m=1$.
If $m>1$, the left <derivative> at $1$ gives $f_Z(s)-s<0$ for $s<1$ sufficiently close to $1$. Since $f_Z(0)\geq0$, <continuity> supplies a <fixed point> below $1$, possibly $0$. Thus \b[survival has positive <probability> precisely when $m>1$, apart from deterministic one-child reproduction].
For the random <bipartite graph>, early exploration of a component alternates between the two <vertex> classes. Before many vertices are exposed, each discovered <vertex> has approximately $\operatorname{Poisson}(c)$ new neighbors in the other class, since $\operatorname{Bin}(n-O(1),c/n)$ tends to that distribution. The two-type mean offspring matrix is
$$
M=\begin{pmatrix}0&c\\c&0\end{pmatrix},
$$
with <spectral radius> $c$; equivalently, the mean growth over two generations is $c^2$. The branching-process comparison therefore predicts a <giant component> exactly when
$$
\boxed{c>1.}
$$
For $c\leq1$, the local process dies out almost surely, so no linear-sized component is expected; $c=1$ is the critical case, not a positive-survival exception. For $c>1$, the symmetric survival <probability> is $1-q>0$, where $q=e^{c(q-1)}$, predicting approximately $2n(1-q)$ vertices in the giant. This last passage from local survival to a unique giant is the usual random-<graph> exploration argument; the branching-process reasoning here supplies the requested threshold explanation rather than a proof of that global theorem.
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