The characteristic disk has one more center than saddle

Write $\beta =h^{\ast }\omega =Pdx+Qdy$ in the oriented source coordinates of the disk and define the characteristic field $X:=(Q,-P)$, so that ${\iota }_X(dx\wedge dy)=\beta$; then $X$ vanishes exactly at the zeros of $\beta$, which are the finitely many nondegenerate interior points $p_1,\dots ,p_N$ of the hypothesis, and $X$ is nowhere zero on $\mathrm{\partial }{D}^2$ and on a collar of it.

Boundary degree. The normalized field $x\mapsto X(x)\mathrm{/}\mathrm{\mid }X(x)\mathrm{\mid }$ is a continuous loop in $S^1$ on the counterclockwise circle $\mathrm{\partial }{D}^2$, and its degree is $1$. If $h(\mathrm{\partial }{D}^2)$ lies in one leaf, then $\beta (\tau )=0$ for the unit tangent $\tau$ of $\mathrm{\partial }{D}^2$, while $\beta \mathrm{\ne }0$ on the boundary; from $0=\beta (\tau )=(dx\wedge dy)(X,\tau )$ the vector $X$ is tangent to the boundary circle and nonzero, hence $X(x)=\epsilon (x)\tau (x)$ with $\epsilon \mathrm{\ne }0$, and $\epsilon$ is continuous on the connected circle, so it has a constant sign: the normalized field is $\pm \tau$ and has the same degree as the unit tangent $\tau$, which is the rotation of the identity map and so has degree $1$ [F5]. If $h{\mathrm{\mid }}_{\mathrm{\partial }{D}^2}$ is a closed transversal, write $X=an+b\tau$ with $n$ the outward unit normal; then $\beta (\tau )=(dx\wedge dy)(X,\tau )=a(dx\wedge dy)(n,\tau )$ with $(dx\wedge dy)(n,\tau )>0$ for the counterclockwise orientation, so $a$ has a fixed nonzero sign, and the straight homotopy $X_s=(1-s)X+s\sigma n$ with $\sigma =\mathrm{sign}a$ has normal component $\sigma ((1-s)\mathrm{\mid }a\mathrm{\mid }+s)\mathrm{\ne }0$, hence is nonzero for all $s\in [0,1]$; the normalized fields are therefore homotopic loops, and the normalized outward normal $n$ has degree $1$ [F5], so the boundary degree is $1$ in both cases.

Outer polygon and holes. Since $\mathrm{\partial }{D}^2$ has a zero-free collar and the zeros are interior, by [F3] we may choose a regular polygon $Q_0$, star-shaped about the origin with positive radial function $r_0(\theta )$, whose boundary lies in the zero-free collar and whose closed convex hull contains all $p_i$. Around each $p_i$ choose pairwise disjoint disks $D_i$ with closures in the interior of $Q_0$ and containing no zero of $X$ other than $p_i$, and inside $D_i$ a centered closed square $Q_i$ with positive radial function ${\rho }_i(\theta )$ about $p_i$. The radial homotopies $\theta \mapsto ((1-s)r_0(\theta )+s)e^{i\theta }$ and $\theta \mapsto p_i+((1-s){\rho }_i(\theta )+s{R}_i)e^{i\theta }$, with $R_i$ the radius of $D_i$, move $\mathrm{\partial }{Q}_0$ to the boundary circle of $D^2$ and $\mathrm{\partial }{Q}_i$ to $\mathrm{\partial }{D}_i$ through loops on which $X$ never vanishes; by the homotopy invariance of the argument increment [F4] the degree of the normalized field on $\mathrm{\partial }{Q}_0$ equals the boundary degree $1$ of step 2.1, and the degree on $\mathrm{\partial }{Q}_i$ equals the degree on $\mathrm{\partial }{D}_i$.

The local degree at a zero is the sign of the Hessian. Fix $p_i$ and work in a foliation chart around $h(p_i)$ with transverse coordinate $z$ and $u=z\circ h$. By [F2] the derivative of $(P,Q)$ at $p_i$ is $(a\circ h)(p_i)D^{2}u(p_i)$, and $X=(Q,-P)$, so $DX(p_i)=(a\circ h)(p_i)J{D}^{2}u(p_i)$ with $J$ the quarter-turn matrix of determinant $1$. The normalized field near $p_i$ is homotopic through nonzero fields on a small circle to the normalized linear field of $DX(p_i)$. If $D^{2}u(p_i)$ is definite, write $H=D^{2}u(p_i)$ and let $\lambda$ be any eigenvalue; the family $H_s=(1-s)H+s\lambda I$ is invertible for every $s\in [0,1]$, so the normalized fields of $J{H}_s$ give a homotopy, and for $H=\lambda I$ the field $JHx$ is, in the complex notation $x_1+i{x}_2$, the map $z\mapsto -\lambda iz$, of degree $1$. If $D^{2}u(p_i)$ is indefinite, choose coordinates diagonalizing it with eigenvalues ${\lambda }_1>0>{\lambda }_2$; the family $(1-s)H+s\mathrm{diag}({\lambda }_1,-{\lambda }_1)$ stays invertible, and for $H=\mathrm{diag}(1,-1)$ one computes $JHx=(-\sin \theta ,-\cos \theta )=-i{e}^{-i\theta }$ on the unit circle, of degree $-1$. Hence a center contributes local degree $+1$ and a saddle contributes $-1$.

The index sum. Cover the closed polygon $Q_0$ by a finite grid of closed axis-parallel rectangles chosen so that every grid line through a side of some square $Q_i$ is a grid line; then each grid cell either lies inside one of the squares $Q_i$ or has interior disjoint from all of them. Discard the cells lying inside a square $Q_i$ and the cells disjoint from $Q_0$; for every remaining cell $C$, the set $C\cap Q_0$ is convex, hence contractible, and is contained in the closed zero-free region $A=Q_0\setminus {\bigcup }_i\mathrm{int}(Q_i)$, so the normalized field $g=X\mathrm{/}\mathrm{\mid }X\mathrm{\mid }$ is defined on $C\cap Q_0$ and the loop $g{\mathrm{\mid }}_{\mathrm{\partial }(C\cap Q_0)}$ extends to a map of the convex set $C\cap Q_0$, hence is nullhomotopic and has argument increment $0$ [F4]. Summing the increments over the finitely many cells, every edge of the grid that lies in the interior of $A$ occurs twice with opposite orientations and cancels by the additivity and reversal rules [F4] (the cells' boundaries are finite polygonal paths, and the common edges are traversed in opposite directions with equal image under $g$); what survives is the boundary of $Q_0$ traversed counterclockwise together with the boundaries of the squares $Q_i$ traversed clockwise. Therefore $0=\mathrm{deg}(g{\mathrm{\mid }}_{\mathrm{\partial }{Q}_0})-{\sum }_i\mathrm{deg}(g{\mathrm{\mid }}_{\mathrm{\partial }{Q}_i})$, that is, $1={\sum }_i{\mathrm{ind}}_{p_i}(X)$.

By step 4.1 the sum ${\sum }_i{\mathrm{ind}}_{p_i}(X)$ equals $c-s$, where $c$ is the number of centers and $s$ the number of saddles among the nondegenerate zeros; step 5.1 gives $c-s=1$, so in particular $c\ge 1$ and the zeros of the characteristic line field are exactly the finitely many nondegenerate centers and saddles. The proof used the relative-genericity supplier, the chain rule, compactness and the elementary degree calculus of circle loops; all of these are choice-free and the only inherited hypothesis is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ of the cooriented interface.