Geometric cardinality is not homotopy invariant

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 ${\mathbb{R}}^2$.