= Solution
At an <equilibrium point of a dynamical system>[equilibrium], the second equation gives $x=y=:s$. The first equation then gives
$$
(a^2-s)(a-s^2)=0.
$$
Hence the fixed-point branches in the $(a,x)$ plane are
$$
\boxed{x=y=a^2\quad(a\in\mathbb R)}
$$
and
$$
\boxed{x=y=\pm\sqrt a\quad(a\geq0).}
$$
The three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt a$, so the second bifurcation value is
$$
\boxed{a^*=1.}
$$
The <Jacobian matrix> at a general point is
$$
J(x,y;a)=
\begin{pmatrix}
y^2-a&-2y(a^2-x)\\
1&-1
\end{pmatrix}.
$$
On the branch $P_0=(a^2,a^2)$ it is triangular, with <eigenvalue>[eigenvalues]
$$
\lambda_1=a^4-a,
\qquad \lambda_2=-1.
$$
Thus $P_0$ is <unstable equilibrium>[unstable] for $a<0$, <asymptotic stability>[asymptotically stable] for $0<a<1$, and unstable for $a>1$.
On a square-root branch write $s=\pm\sqrt a$. Then
$$
J(P_s)=
\begin{pmatrix}
0&2s^2(1-s^3)\\
1&-1
\end{pmatrix},
$$
whose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt a$ is therefore a <saddle equilibrium> for every $a>0$. The positive branch is a <saddle equilibrium>[saddle] for $0<a<1$ and <asymptotic stability>[asymptotically stable] for $a>1$; close to $a=1$ it is a <stable node>.
To resolve the nonhyperbolic point at $a=0$, make the prescribed substitution
$$
X=x-a^2,
\qquad Y=y-a^2
$$
and append $\dot a=0$. The equations become
$$
\dot X=-aX+X(Y+a^2)^2,
\qquad
\dot Y=X-Y,
\qquad
\dot a=0.
$$
The centre subspace is $Y=X$. Seek the <extended centre manifold for a parameter> as
$$
Y=h(X,a)
=X+AX^2+BXa+Ca^2+O(3).
$$
The <centre-manifold invariance equation> is
$$
h_X\dot X=X-h.
$$
To second order, $\dot X=-aX+O(3)$, so comparison of coefficients gives
$$
-aX=-AX^2-BXa-Ca^2,
$$
and hence $A=C=0$, $B=1$. Therefore
$$
\boxed{Y=X+aX+O(3).}
$$
Substitution into the $X$ equation gives the reduced flow
$$
\boxed{\dot X=-aX+X^3+O(4).}
$$
Its central branch $X=0$ is <unstable equilibrium>[unstable] for $a<0$ and <asymptotic stability>[stable] for $a>0$, while the two nonzero branches $X\sim\pm\sqrt a$ for $a>0$ are <unstable equilibrium>[unstable]. Thus the bifurcation at zero is a <subcritical pitchfork bifurcation> with reversed normal-form parameter $\mu=-a$, exactly as recorded by the <extended centre manifold of the 2023 Cambridge quadratic-product system>.
The complete <bifurcation diagram> is therefore as follows. For $a<0$, only $P_0$ exists and is <unstable equilibrium>[unstable]. For $0<a<1$, $P_0$ is <asymptotic stability>[stable] while both $P_+$ and $P_-$ are saddles. For $a>1$, $P_+$ is <asymptotic stability>[stable] while $P_0$ and $P_-$ are saddles. At $a=1$, $P_0$ and $P_+$ cross and exchange stability, so the bifurcation is <transcritical bifurcation>[transcritical]. This gives the <bifurcation diagram of the 2023 Cambridge quadratic-product system>.
Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt a,\sqrt a)$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt a,\sqrt a)$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional <stable manifold>[stable and unstable manifolds]; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.
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