= 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 $g$ is
$$
r(\mu)=\frac{-1+\sqrt{1+4\mu}}2=\mu+O(\mu^2),\qquad g'(r)=-2r.
$$
For 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 $\mu\uparrow0$ these <fixed points> collide at zero. For small positive $\mu$, $r>0$ and
$$
f_L(r)=-r,\qquad f_L(-r)=r,
$$
so a stable symmetric two-cycle is present, with <periodic-orbit multiplier> $4r^2<1$. Hence the global transition is
$$
\boxed{\mu<0:\text{ two stable fixed points};\quad\mu=0:\text{ the zero fixed point};\quad
\mu>0:\text{ one stable two-cycle}.}
$$
The 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-85-lorenz-maps.png]
{title=The two Lorenz-map branches and their even quadratic companion as the parameter crosses zero}
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