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Assume the $g$ cycle has least period $n$ and nonzero periodic-orbit multiplier $P=(g^n{)}^{\prime }(\hat{x})$. It then avoids zero. Combining the iterate identity with the periodic-orbit multiplier-sign formula givesThus $P>0$ makes $\hat{x}$ return after $n$ iterates, whereas $P<0$ gives $f_L^n(\hat{x})=-\hat{x}$. In the latter case oddness on the noncritical orbit impliesTo prove the least periods, note that a $g$ cycle cannot contain two distinct points $x$ and $-x$: evenness would give them the same image, but $g$ permutes its distinct cycle points bijectively. Therefore the $n$ absolute values in the cycle are distinct. Since $|f_L^k(\hat{x})|=|g^k(\hat{x})|$, any return under $f_L$ must occur at a multiple of $n$. For $P<0$ the first such return fails because $\hat{x}\ne 0$, so $2n$ is minimal. Hence the period transfer from a quadratic map to its Lorenz sign lift isThere are two symmetry-related $f_L$ cycles in the positive case and one cycle containing both signs in the negative case. Away from zero $f_L^{\prime }(x)=2|x|$, so their periodic-orbit multipliers are $|P|$ and $|P{|}^2$, respectively. A stable $g$ cycle therefore gives stable lifted cycles in both cases.

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 givesThe second factor yields a two-cycleIt exists for $\mu >3/4$ and has periodic-orbit multiplier $4pq=4(1-\mu )$. Its stability range is $3/4<\mu <5/4$, so $g$ 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 iterateThe one-sided limits at zero are opposite, and agree with zero exactly when $\mu =1$ in this neighborhood. There $F_1(x)=\mathrm{sgn}(x)(2x^2-x^4)$ 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.

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 cyclesFor $1<\mu <5/4$, $p>0$ and $q<0$, so the periodic-orbit multiplier of the $g$ cycle is negative and the lifted orbit isIt is one stable four-cycle. Hence the gluing of cycles in an iterated Lorenz map givesEquivalently, $F_{\mu }$ 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 ${\lim }_{x\to 1^{-}}{F}_1(x)=1$ and ${\lim }_{x\to 1^{+}}{F}_1(x)=-1$. 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.