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The origin's Jacobian matrix is triangular, $J=\left(\begin{array}{cc}a+1& 1\\ 0& b-1\end{array}\right)$, so its Floquet multipliers are $a+1$, $b-1$. It is attracting for $-2<a<0$, $0<b<2$. The candidate codimension-one boundaries are $a=0$, $a=-2$, $b=2$, $b=0$: respectively a $+1$ or $-1$ Floquet multiplier. No complex-pair crossing is possible at the origin.

The map is odd. To locate the nearby steady branches, write $S=x^2>0$. The first fixed-point equation gives $y=x(S-a)$; substituting into the second givesFor a symmetric two-cycle solve $F(x,y)=-(x,y)$. Then $y=x(S-a-2)$ andThese exact orbit equations for a triangular cubic map determine the amplitude and parameter side without confusing the original cubic coefficients with their centre manifold values.

On $a=0$, with $b\ne 2$, steady states have $x^2=a(2-b)+o(a)$ and $y=-ax+x^3$. The pitchfork bifurcation is supercritical for $b<2$ and subcritical for $b>2$. Its centre manifold cubic coefficient is $-1/(2-b)$. In the stable-origin boundary segment $0<b<2$, an attracting pair appears on $a>0$.

On $a=-2$, with $b\ne 0$, the two-cycle has $x^2=b(a+2)+o(a+2)$. The centre manifold map has cubic coefficient $-1/b$ and Floquet multiplier $-1+(a+2)$. Since its unstable side is $a+2<0$, the period-doubling bifurcation is supercritical for $b<0$ and subcritical for $b>0$. In particular the boundary segment $0<b<2$ is subcritical: the nearby two-cycle lies on $a>-2$, where the origin is still attracting, and is unstable in the critical direction.

On $b=2$, with $a\ne 0$, the steady branch hasIts scalar centre manifold map is $y^{\prime }=(1+b-2)y+(1+a^{-3})y^3+\cdots$. It is supercritical for $-1<a<0$, and subcritical for $a<-1$ or $a>0$. At $a=-1$ the cubic vanishes: the exact equation gives $b-2=-3S^2+O(S^3)$, so this is a degenerate pitchfork bifurcation with destabilizing quintic coefficient and subcritical quartic amplitude scaling. It is not a generic point of the codimension-one curve.

On $b=0$, with $a\ne -2$, put $A=a+2$. The two-cycle satisfiesThe centre manifold map's cubic coefficient is $1-A^{-3}$. Thus the period-doubling bifurcation is subcritical for $-2<a<-1$, and supercritical for $a<-2$ or $a>-1$. In the stable-origin edge $-1<a<0$, a stable small two-cycle appears for $b<0$. At $a=-1$ the cubic vanishes, requiring the calculation below. All criticality labels concern the centre manifold direction; a bifurcating orbit is fully attracting only if the remaining Floquet multiplier is also stable. The sketch marks the small branches on the appropriate sides of each boundary of the stable-origin rectangle.

For clarity, these stability labels follow directly from one-dimensional derivatives. At a pitchfork bifurcation $z^{\prime }=(1+\eta )z+c{z}^3$, the nonzero state has $z^2=-\eta /c$ and Floquet multiplier $1-2\eta$ to first order. At a period-doubling bifurcation $z^{\prime }=-(1+\eta )z+c{z}^3$, the symmetric cycle has $z^2=\eta /c$ and one-step derivative $-1+2\eta$; its two-step Floquet multiplier is $1-4\eta +o(\eta )$. A supercritical branch lies on $\eta >0$ and is attracting in the centre manifold direction; a subcritical branch lies on $\eta <0$ and is repelling there.

At $(a,b)=(-1,0)$ the Floquet multipliers are $0,-1$, so the degeneracy is not two simultaneous unit Floquet multipliers. On its centre manifold, $x=-y-y^3+O(y^7)$ and direct substitution gives $y^{\prime }=-y-3y^5+O(y^7)$. With $\alpha =a+1$ small, the unfolding to leading weighted order isThis is a generalized flip bifurcation: one parameter crosses $-1$, and the other changes the cubic sign. Writing $S\simeq y^2$, its two-cycles obey $b+3\alpha S-3S^2=0$. For $\alpha >0$ a stable smaller two-cycle and an unstable larger two-cycle collide atThe intervening range $-3{\alpha }^2/4<b<0$ contains both. For $\alpha <0$ the period-doubling bifurcation is subcritical. At $\alpha =0$, the exact orbit equation is $3S^2-3S^3+S^4=b$, so the unstable two-cycle exists for $b>0$ with $|y|\sim (b/3{)}^{1/4}$. This proves both the codimension and the required quintic effect.