Latitude and meridian intersections on the torus

By [F2] the sets $A$ and $B$ are embedded circles, and $A\cap B=\{([0],[0])\}$: a point of $A$ has second coordinate $[0]$, a point of $B$ has first coordinate $[0]$. At that point $T_{p}A=\mathbb{R}{\mathrm{\partial }}_x$ and $T_{p}B=\mathbb{R}{\mathrm{\partial }}_y$, and $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ is the positive product frame, so $A$ and $B$ are transverse and the local sign is $+1$ by [F3]. Hence $I(A,B)=1$; with the factors exchanged the ordered basis $({\mathrm{\partial }}_y,{\mathrm{\partial }}_x)$ is negative, so $I(B,A)=-1$.

The same computation counts mod two: the transverse intersection has one point, so $I_2(A,B)=I_2(B,A)=1$ by [F4], and no orientation hypothesis was used. For the perturbation, each $B_t:=\{[t]\}\times Q$ with $t\in \mathbb{R}$ is again a meridian circle with $T_p{B}_t=\mathbb{R}{\mathrm{\partial }}_y$ at its unique intersection point $([t],[0])$ with $A$, so the intersection is transverse with one point and $I_2(A,B_t)=1$ for every $t$; the explicit family $t\mapsto \{[t]\}\times Q$ thus has constant mod 2 count by this direct computation. Thus the example realizes the source's count $I_2=1$ and the sign asymmetry $I(A,B)=-I(B,A)$; no choice principle is used: the circles, the product structure and the perturbation are explicit, the perturbation count is computed directly for every $t$, without selecting a homotopy or applying classification.