= Solution
Write $D=AB-1>0$. The physically nonnegative branches, omitting $S_1$, are
$$
\boxed{S_2=(1,0),\qquad S_3=(0,\gamma),\qquad S_4=\left(\frac{A\gamma-1}{D},\frac{B-\gamma}{D}\right)\quad(1/A\leq\gamma\leq B)}.
$$
On both <bifurcation diagrams>, draw $S_2$ solid for $\gamma<B$ and dashed for $\gamma>B$; draw $S_3$ dashed for $\gamma<1/A$ and solid for $\gamma>1/A$. The coexistence branch $S_4$ is dashed throughout its interior because it is a <saddle point>. It connects $S_3$ at $\gamma=1/A$ to $S_2$ at $\gamma=B$, through two <transcritical bifurcations> when the branches are continued mathematically beyond the nonnegative quadrant.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2023-squirrel-bifurcations.png]
{title=Stable boundary populations and the unstable coexistence threshold}
{description=Original <bifurcation diagrams> for $A=2$, $B=0.75$. Solid branches are asymptotically stable, dashed branches unstable. The physically admissible coexistence branch joins the grey-only branch at $\gamma=0.5$ to the red-only branch at $\gamma=0.75$.}
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