Oriented and Mod Two Intersection Numbers — Examples

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.

Distinct great circles $C_1,C_2$ are planes through the origin meeting the sphere, and distinct planes through the origin in ${\mathbb{R}}^3$ meet in a line through the origin, which cuts $S^2$ in exactly two antipodal points; at such a point $p$ the tangent line $T_p{C}_i=p^{\perp }\cap P_i$ lies in $T_p{S}^2=p^{\perp }$ and determines the plane $P_i$ as $\mathrm{span}\{p,T_p{C}_i\}$, so the two tangent lines are distinct one-dimensional subspaces of the two-dimensional $T_p{S}^2$ and therefore span it, which is transversality.

Each great circle is antipodally invariant. Its antipodal quotient is a circle, and the induced map $C_i\mathrm{/}(\pm 1)\to {\mathbb{R}\mathbb{P}}^2$ is injective. Since its source is compact and its target Hausdorff, it is a homeomorphism onto its image; in the local diffeomorphism charts of [F2] it is the embedded arc $C_i$, hence is a smooth embedding. Saturation of the two $C_i$ under the antipodal involution gives $q(C_1)\cap q(C_2)=q(C_1\cap C_2)$, a single point. The invertible differential of $q$ carries the two distinct tangent lines of 1.1 to distinct tangent lines in the quotient, preserving transversality.

By [F3] the projective plane is nonorientable, so no oriented intersection number of the two lines is defined. By [F4] the mod 2 intersection number is defined and equals the parity of the intersection, namely $1$; under the stated ${\mathrm{A}\mathrm{C}}_{\omega }$, every homotopic pair of transverse representatives of the two projective lines meets an odd number of times, so the lines can never be deformed to disjoint positions.

Regularity makes $F$ transverse to $\{y\}$. Its finite signed count is the sum in [F1], since the local intersection signs against the positive point equal $\mathrm{sgn}(d{F}_p)$ by [F2]. The sum is $\mathrm{deg}(F)$; with compact $M$ the definition in [F2] names it $I(F,\{y\})$. Properness supplies finiteness even when $M$ is noncompact, but it does not enlarge that compact-source definition.

Assume $M$ compact. At $(p,y)$ a fibre tangent vector is $(u,0)$ and a graph tangent vector is $(v,d{F}_{p}v)$. The ordered derivative matrix for fibre first, graph second is $\left(\begin{array}{cc}I& I\\ 0& d{F}_p\end{array}\right)$, whose determinant has sign $\mathrm{sgn}(d{F}_p)$ in the induced orientations. The determinant-line calculation also handles $n=0$: the two source point signs from $M$ cancel, leaving the ambient point sign of $M\times N$, equal to $\mathrm{sgn}(d{F}_p)$. The transverse intersection is exactly the finite regular fibre, so summing gives $I(M\times \{y\},{\mathrm{\Gamma }}_F)=\mathrm{deg}(F)$. Reversing the two $n$-blocks multiplies each sign by $(-1{)}^{n^2}=(-1{)}^n$.

Write $s(p)=(p,F(p))$ and let $i$ include $M\times \{y\}$. Their oriented parametrizations identify $I(s,i)$ with the graph-first count, so [F4] gives $I(s,i)=(-1{)}^{n}I(s\times i,{\mathrm{\Delta }}_{M\times N})$. This agrees with the opposite-order graph sign in 2.1, establishing compatibility of the degree and diagonal conventions.

Write $\theta =2\pi u$ modulo $2\pi$. For $(\cos \theta ,s+\sin \theta )\in A$ one needs $\sin \theta =-s$. For $0\le s<1$ there are exactly two solutions, one with $\cos \theta >0$ and one with $\cos \theta <0$; for $s=1$ there is a single solution $[u]=[3\mathrm{/}4]$, a tangency; for $s>1$ there is none. In particular the raw cardinality of $f_s^{-1}(A)$ is $2$ for $0\le s<1$ and $0$ for $s>1$.

At a solution the ordered pair $(d{f}_{\theta }(1),(1,0))=((-\sin \theta ,\cos \theta ),(1,0))$ has determinant $-\cos \theta$ in the standard basis, so the local sign is $\epsilon =-\mathrm{sgn}(\cos \theta )$ by [F2]. Hence at the two solutions of 1.1 the signs are $+1$ and $-1$, and both the signed count $I(f_s,A)$ and the parity count $I_2(f_s,A)$ are $0$ for every transverse slice; for $s>1$ the fibre is empty and both counts are again $0$.

The family $f_s$ is a smooth homotopy between $f_0$ and $f_2$, yet the raw cardinalities of the intersections are $2$ and $0$; hence geometric cardinality is not a homotopy invariant. The signed and parity counts, by contrast, are constant with value $0$ across the family and are compatible with the invariance asserted under the hypotheses of [F5]; the cardinality changes exactly at the tangency $s=1$, where the slice is not transverse. The full evaluation map remains transverse since its derivative in $s$ is $(0,1)$, which together with $TA$ spans ${\mathbb{R}}^2$.

For $t<1$, $F_t^{-1}(0)=\{0,1\mathrm{/}(1-t)\}$ and ${\mathrm{\partial }}_x{F}_t$ is $+1$ at $0$ and $-1$ at $1\mathrm{/}(1-t)$. At $t=1$, $F_1(x)=x$ has the single zero $0$ with derivative $+1$. Every slice is transverse at its zeros. The cardinalities change from $2$ to $1$, the signed counts from $0$ to $1$, and the parity counts from $0$ to $1$ as $t$ reaches $1$.

Every slice is proper. For $t<1$, $\mathrm{\mid }x-(1-t)x^2\mathrm{\mid }\to \mathrm{\infty }$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$; for $t=1$, the map is the identity. Thus each preimage of a compact real set is closed and bounded, hence compact. The trace contains $\{(1\mathrm{/}(1-t),t):0\le t<1\}$, an unbounded branch whose intersection point escapes to $+\mathrm{\infty }$ as $t\uparrow 1$. Consequently $F^{-1}(\{0\})$ is noncompact and the combined map is not proper, although both endpoint maps, indeed all slices, are proper. This satisfies the refuted claim's hypotheses and disproves its conclusion.