= Solution
A <fixed point> $X_*$ satisfies $F(X_*)=X_*$. The <Jacobian matrix> is $DF=\bigl(\begin{smallmatrix}f_x&f_y\\g_x&g_y\end{smallmatrix}\bigr)$; 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 $(1-a-b)x=1$. Therefore
$$
\boxed{X_+=\frac{(1,b)}{1+a-b}\quad\text{when }b<1+a,\qquad
X_-=\frac{(1,b)}{1-a-b}\quad\text{when }b>1-a.}
$$
There 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>.
\Image[/past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2007-lozi-existence.png]
{title=Lozi fixed-point existence regions: positive and negative branches lie in distinct half-planes that can overlap}
For $x\ne0$, $DF=\bigl(\begin{smallmatrix}-a\operatorname{sgn}x&1\\b&0\end{smallmatrix}\bigr)$. 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, $DF^2=-I$ 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 $0\leq x\leq1$ is mapped by $x\mapsto1-x$, 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 $a\leq-1$ 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 $hx_2=1-a|x_1|$, $hx_1=1-a|x_2|$. 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 $\boxed{x_1x_2<0}$.
Choose $x_1>0>x_2$. Solving $ax_1+hx_2=1$, $hx_1-ax_2=1$ gives
$$
\boxed{x_1=\frac{a+h}{a^2+h^2},\qquad x_2=\frac{h-a}{a^2+h^2},\qquad
X_1=(x_1,bx_2),\quad X_2=(x_2,bx_1).}
$$
The 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>.
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