= Solution
Under $A>1$, $B<1$, $AB>1$, the thresholds are ordered $0<1/A<B<1$. As $\gamma$ decreases from one, $S_2$ first becomes stable at
$$
\boxed{\gamma_\mathrm{local}=B=\alpha_{21}}.
$$
For $1/A<\gamma<B$, both boundary equilibria are stable, and the positive coexistence point $S_4$ is a <saddle point>. Its stable manifold is the separatrix between their <basins of attraction>. For every fixed $u_1(0)>0$, a sufficiently small positive $u_2(0)$ lies in the red-survival basin. To justify the quantifier beyond a neighborhood of $S_2$, start on the red-only axis: its <logistic equation> approaches $S_2$. Continuous dependence over the finite time needed to enter a stable neighborhood shows that sufficiently small positive initial grey populations do so as well.
At the second threshold,
$$
\boxed{\gamma_\mathrm{global}=1/A=1/\alpha_{12}},
$$
$S_4$ meets $S_3$. For $0<\gamma<1/A$, $S_3$ is a <saddle point>, its stable manifold is the grey-only axis, and there is no positive coexistence <equilibrium>. Boundedness and the same <Bendixson-Dulac theorem> argument rule out an interior cycle. Consequently every initial state with $u_1(0)>0$ and $u_2(0)\geq0$ approaches $S_2=(1,0)$. This is the removal regime that eliminates grey squirrels regardless of their initial abundance. At threshold equality linearization is nonhyperbolic; the strict parameter intervals above avoid inferring stability from a zero <eigenvalue>.
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