= Solution
Take small incoming sections $y=\pm h$ and outgoing sections $x=\pm h$ after straightening the <stable manifold> and <unstable manifold> of the origin. Its <eigenvalues> are $1,-\delta$, so the leading local passage satisfies $x(t)=x_0e^t$, $y(t)=y_0e^{-\delta t}$. The exit time and transverse exit coordinate are
$$
T=\log\frac{h}{|x_0|},\qquad y_{\mathrm{out}}=\operatorname{sgn}(y_0)h^{1-\delta}|x_0|^\delta.
$$
Thus the local passage contracts strongly as $x_0\to0$, while its duration diverges. Smooth global reinjection from the right outgoing section to the upper incoming section has leading transverse coordinate $-\mu+K_Ry_{\mathrm{out}}$; the corresponding left-to-lower return has $\nu+K_Ly_{\mathrm{out}}$. Choose the parameters as these signed splitting coordinates and absorb $h^{1-\delta}$ into $A,B$.
After scaling the incoming-section labels to $y_n=\pm1$, the leading <Poincaré return map> is
$$
x_{n+1}=\begin{cases}-\mu+A\operatorname{sgn}(y_n)|x_n|^\delta,&x_n>0,\\
\nu+B\operatorname{sgn}(y_n)|x_n|^\delta,&x_n<0,
\end{cases}\qquad y_{n+1}=\operatorname{sgn}(x_n).
$$
The second coordinate records which outgoing branch was used, not an independent continuous amplitude. A smooth planar flow preserves the orientation of a transverse-section return. For the right-to-upper pair of sections, the ordered pair consisting of the flow direction and the positive section tangent has the same orientation at both ends; the same is true for the left-to-lower pair. Hence $A,B>0$ for these sign conventions and a nondegenerate global return.
For general nonlinear $P,Q$, this is an \b[asymptotic return model], with smooth parameter corrections and higher-order local-passage corrections. The hypotheses do not imply an exact power-law map in the original coordinates. Interpreting the printed parameters themselves as linear splitting coordinates also requires a nondegenerate two-parameter unfolding: the two splitting derivatives must be nonzero. For example, composing a generic right-loop splitting with $\mu\mapsto\mu^2$ preserves the stated loop at $\mu=0$ but does not give the printed signed $-\mu$ offset. The subsequent region diagram uses the standard generic splitting-coordinate interpretation. Nor is $x=0$ a normal return point: it lies on the <stable manifold>, whose trajectory takes infinite time to reach the <saddle equilibrium>. This explains how the map describes nearby recurrent trajectories without extending a finite-time return across the singular separatrix.
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