= Solution
The origin's <Jacobian matrix> is triangular, $J=\begin{pmatrix}a+1&1\\0&b-1\end{pmatrix}$, 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 gives
$$
\boxed{S[1-(S-a)^3]=a(2-b).}
$$
For a symmetric <two-cycle> solve $F(x,y)=-(x,y)$. Then $y=x(S-a-2)$ and
$$
\boxed{S[1+(S-a-2)^3]=b(a+2).}
$$
These 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 $b\ne2$, 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 $b\ne0$, 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 $a\ne0$, the steady branch has
$$
y^2\sim-\frac{a^3}{1+a^3}(b-2).
$$
Its scalar <centre manifold> map is $y'=(1+b-2)y+(1+a^{-3})y^3+\cdots$. 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> satisfies
$$
y^2\sim\frac{A^3}{1-A^3}b.
$$
The <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> $z'=(1+\eta)z+cz^3$, the nonzero state has $z^2=-\eta/c$ and <Floquet multiplier> $1-2\eta$ to first order. At a <period-doubling bifurcation> $z'=-(1+\eta)z+cz^3$, 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 $y'=-y-3y^5+O(y^7)$. With $\alpha=a+1$ small, the unfolding to leading weighted order is
$$
\boxed{y'=(-1+b)y+3\alpha y^3-3y^5+\cdots.}
$$
This 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 at
$$
\boxed{b=-\tfrac34\alpha^2+o(\alpha^2),\qquad S=\tfrac12\alpha+o(\alpha).}
$$
The 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.
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