 Solution

The associated even map is simply $g(x)=\mu -x^2$. For nonzero $x$, the odd quadratic Lorenz map has the two branches $f_L(x)=x^2-\mu$ on the positive side and $f_L(x)=\mu -x^2$ on the negative side. Their limiting values at zero are $-\mu$ and $+\mu$, while the prescribed point value is $f_L(0)=-\mu$. Thus the branches join continuously only at $\mu =0$, where the origin is a fixed point and $f_L(x)=x|x|$.

The small fixed point of $g$ isFor small negative $\mu$, $r<0$ and the odd quadratic Lorenz map has two stable fixed points $r,-r$. Their periodic-orbit multiplier is $2|r|<1$. As $\mu \uparrow 0$ these fixed points collide at zero. For small positive $\mu$, $r>0$ andso a stable symmetric two-cycle is present, with periodic-orbit multiplier $4r^2<1$. Hence the global transition isThe large fixed points $\pm (1+r)$ remain unstable near this transition and are not created or destroyed there. The parabola $g$ itself is smooth as $\mu$ passes zero; its fixed point crosses the critical point and changes the sign of its periodic-orbit multiplier. The change of orbit type occurs in the sign-lifted discrete dynamical system, whose discontinuity switches orientation.