= Solution
The positive <fixed point> $r=(-1+\sqrt{1+4\mu})/2$ of $g$ has <periodic-orbit multiplier> $-2r$. At $\mu=3/4$, $r=1/2$ and the <periodic-orbit multiplier> crosses $-1$. Factoring the second-iterate fixed-point equation gives
$$
g^2(x)-x=-(x^2+x-\mu)(x^2-x+1-\mu).
$$
The second factor yields a two-cycle
$$
p=\frac{1+\sqrt{4\mu-3}}2,\qquad q=\frac{1-\sqrt{4\mu-3}}2,\qquad
g(p)=q,\quad g(q)=p.
$$
It exists for $\mu>3/4$ and has <periodic-orbit multiplier> $4pq=4(1-\mu)$. Its stability range is $3/4<\mu<5/4$, so \b[$g$ undergoes a supercritical period-doubling <bifurcation> at $\mu=3/4$]. At $\mu=1$ the two-cycle is $\{0,1\}$ and its <periodic-orbit multiplier> is zero, making it superstable.
For positive $\mu$ and $0<|x|<\sqrt\mu$, direct composition gives the local second iterate
$$
F_\mu(x)=f_L^2(x)=\operatorname{sgn}(x)\bigl(\mu-\mu^2+2\mu x^2-x^4\bigr),\qquad
F_\mu(0)=\mu-\mu^2.
$$
The one-sided limits at zero are opposite, and agree with zero exactly when $\mu=1$ in this neighborhood. There $F_1(x)=\operatorname{sgn}(x)(2x^2-x^4)$ is continuous at zero and fixes it. Thus the local return branch has the same discontinuity-closing transition as the map in part (a), with effective parameter $\mu^2-\mu$ crossing zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-85-lorenz-second-iterate.png]
{title=Local second-iterate Lorenz return maps near zero, showing the discontinuity close and reverse at parameter one}
For $3/4<\mu<1$, both $p,q$ are positive and the <period transfer from a quadratic map to its Lorenz sign lift> gives the two stable cycles
$$
\{p,-q\},\qquad\{q,-p\}.
$$
For $1<\mu<5/4$, $p>0$ and $q<0$, so the <periodic-orbit multiplier> of the $g$ cycle is negative and the lifted orbit is
$$
p\longmapsto-q\longmapsto-p\longmapsto q\longmapsto p.
$$
It is one stable four-cycle. Hence the <gluing of cycles in an iterated Lorenz map> gives
$$
\boxed{\mu<1:\text{ two stable two-cycles of }f_L;\qquad
\mu>1:\text{ one stable four-cycle of }f_L.}
$$
Equivalently, $F_\mu$ changes from two small stable <fixed points> $\pm q$ to a stable small two-cycle $\pm(-q)$. At the critical value itself, the actual convention gives $f_L(0)=-1$, $f_L(-1)=0$, $f_L(1)=0$: the surviving critical two-cycle is $\{0,-1\}$, and $+1$ is preperiodic. This critical itinerary must be evaluated directly, not with part (b)'s noncritical sign product.
There is also a precise qualification to the printed use of “continuous”. The whole second iterate is not continuous everywhere at $\mu=1$: $F_1(1)=-1$, while $\lim_{x\to1^-}F_1(x)=1$ and $\lim_{x\to1^+}F_1(x)=-1$. Jumps persist at the nonzero preimages of the original discontinuity. Therefore the stated global-<bifurcation> criterion holds for the return map restricted to a neighborhood of zero, as in the sketches; if continuity on all of $\mathbb R$ is demanded literally, the claim about $f_L^2$ needs that restriction. The orbit creation and destruction above is unaffected.
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