A fixed point $X_{\ast }$ satisfies $F(X_{\ast })=X_{\ast }$. The Jacobian matrix is $DF=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)$; its eigenvalues are the linear fixed-point multipliers used in (i)–(iii).

For the Lozi map, $y=bx$ and $x=1+bx-a|x|$. At a positive point this gives $(1+a-b)x=1$, and at a negative point $(1-a-b)x=1$. ThereforeThere is no fixed point at $x=0$, because then $y=0$ and the first component would be one. On an equality boundary the corresponding scalar equation is $0=1$, so that branch is absent, not a line of fixed points.

For $x\ne 0$, $DF=\left(\begin{array}{cc}-a\mathrm{sgn}x& 1\\ b& 0\end{array}\right)$. If $a=0$, there is exactly one fixed point $(1,b)/(1-b)$ unless $b=1$, when there is none. Its eigenvalues satisfy ${\lambda }^2=b$. It is a sink fixed point of a map for $|b|<1$ and a source fixed point of a map for $|b|>1$. At $b=-1$, the affine map about its fixed point is a quarter turn, $D{F}^2=-I$ and $F^4=I$, so it is neutral, not attracting. At $b=1$ no fixed point exists. The positive/negative branch labels follow the sign of $1-b$.

If $b=0$, $X_{+}=(1/(1+a),0)$ exists for $a>-1$ with fixed-point multipliers $-a,0$. It is a sink fixed point of a map for $-1<a<1$ and a forward saddle fixed point of a map for $a>1$. At $a=1$ the fixed-point multiplier $-1$ is neutral; on the invariant axis the interval $0\le x\le 1$ is mapped by $x\mapsto 1-x$, giving two-cycles and the fixed midpoint. The negative point $X_{-}=(1/(1-a),0)$ exists only for $a>1$ and has fixed-point multipliers $a,0$, again a forward saddle fixed point of a map. For $a\le -1$ there is no fixed point. At $a=b=0$, both fixed-point multipliers vanish and two iterates send all initial points to the fixed point.

For a genuine two-cycle let the first coordinates be $x_1,x_2$ and put $h=1-b$. The two equations are $h{x}_2=1-a|x_1|$, $h{x}_1=1-a|x_2|$. If both first coordinates had the same sign $s$, subtracting would give $(h-as)(x_2-x_1)=0$. Since distinct cycle points must have different first coordinates, this contradicts $a^2\ne h^2$. If one coordinate were zero, the remaining equations would instead force $a=\pm h$, also excluded. Hence $\boxed{x_1{x}_2<0}$.

Choose $x_1>0>x_2$. Solving $a{x}_1+h{x}_2=1$, $h{x}_1-a{x}_2=1$ givesThe required signs hold exactly when $\boxed{a>|1-b|}$. This is the unique two-cycle up to interchanging its two points, under the stipulated nondegeneracy. Borderline parameter lines can carry different degenerate families and were explicitly excluded. These are Lozi fixed points and the alternating-sign two-cycle.

The equilibria are $\boxed{(0,0),(\pm \sqrt{2/3},0)}$. The Jacobian is $J=\left(\begin{array}{cc}0& 1\\ 2-9x^2& -1\end{array}\right)$. At the origin it has eigenvalues $1,-2$ and tangent eigenvectors $(1,1)$ and $(1,-2)$, so the origin is a saddle equilibrium. At either nonzero point the eigenvalues are $\boxed{(-1\pm i\sqrt{15})/2}$, giving stable foci.

The energy-like function $H=y^2/2-x^2+3x^4/4$ has $\dot{H}=-y^2\le 0$. It is coercive at large distance, so trajectories remain bounded; there are no periodic orbits because their energy could not return to its starting value unless $y\equiv 0$. The stable manifold of the saddle equilibrium separates the two attracting basins. Its unstable branches leave the origin and approach the right and left foci, respectively. Their negative energy near the origin prevents crossing back through $x=0$.

For either invariant manifold write $y=h(x)=sx+c{x}^3+O(x^5)$. Odd symmetry rules out even powers. Tangency to the vector field requires $h^{\prime }(x)h(x)=2x-h(x)-3x^3$. Matching coefficients yields $s^2+s-2=0$ and $c(4s+1)=-3$. HenceThese are local graphs, not globally exact cubic curves.

The Lorenz equilibria satisfy $y=x$, $z=x^2/b$ and $x(r-1-z)=0$. ThusAt the origin the third eigenvalue is $-b$ and the other two solve ${\lambda }^2+(\sigma +1)\lambda +\sigma (1-r)=0$. Hence the origin is stable for $r<1$, has one zero eigenvalue at $r=1$, and becomes a saddle equilibrium with one unstable direction for $r>1$. The symmetry $(x,y,z)\mapsto (-x,-y,z)$ interchanges the new branches.

The nonzero-equilibrium characteristic polynomial isNear $r=1$ its Routh-Hurwitz inequalities are strict, so both new branches are stable. More explicitly, center-manifold projection at $r=1$ givesThe linear coefficient crosses positively and the cubic coefficient is negative. Therefore the equilibrium bifurcation is a supercritical pitchfork at $r=1$, not a saddle equilibrium-node.

For completeness, the later stability boundary satisfiesIf $\sigma \le b+1$, the nonzero branches remain linearly stable for all $r>1$. If $\sigma >b+1$, they lose stability atAt that point $P=(\lambda +\sigma +b+1)({\lambda }^2+{\omega }_H^2)$ andThis establishes a genuine transverse Hopf crossing. It is subcritical for these positive parameters, so it is not the supercritical Hopf normal form of a dynamical system from (c). One way to check its nonlinear sign is the cubic Hopf formula. At $C_{+}$ write $s=\sqrt{b(r_H-1)}$ andFor right and adjoint eigenvectors $Aq=i{\omega }_{H}q$, $A^{\ast }p=-i{\omega }_{H}p$, $⟨p,q⟩=1$, the cubic coefficient has the sign ofSubstitution and clearing positive factors reduces its sign to that of the following polynomial in $h=\sigma -b-1>0$:Every term is positive. The radial cubic is therefore positive, giving the unstable small cycle on the stable-equilibrium side. There is no saddle equilibrium-node among the equilibrium branches for $\sigma ,b,r>0$: the only zero-eigenvalue branch intersection is the symmetry-forced pitchfork. This does not rule out separate global bifurcations of periodic orbits.