The Mobius band presented by two regular patches, with normal comparison on the interiors of the overlap components

Writing $c=\cos (t\mathrm{/}2)$ and $q=\sin (t\mathrm{/}2)$, the identities in [L1], [L2], and [L3] give $\mathrm{\Psi }(t+2\pi ,s)=\mathrm{\Psi }(t,-s)$ for every $(t,s)$, because $\cos ((t+2\pi )\mathrm{/}2)=-c$ and $\sin ((t+2\pi )\mathrm{/}2)=-q$ while $\cos (t+2\pi )=\cos t$ and $\sin (t+2\pi )=\sin t$.

The two rectangles $R_1$ and $R_2$ each have angle length $4\pi \mathrm{/}3<2\pi$, their union covers a full turn modulo $2\pi$, they overlap directly on $K_1=[5\pi \mathrm{/}6,7\pi \mathrm{/}6]\times [-1\mathrm{/}2,1\mathrm{/}2]$, and after the shift in step 1.1 they overlap again through $K_2=[-\pi \mathrm{/}6,\pi \mathrm{/}6]\times [-1\mathrm{/}2,1\mathrm{/}2]$ on $R_1$ against $[11\pi \mathrm{/}6,13\pi \mathrm{/}6]\times [-1\mathrm{/}2,1\mathrm{/}2]$ on $R_2$.

Differentiating and using [F3] and [L1] gives ${\mathrm{\Psi }}_t\times {\mathrm{\Psi }}_s=(1+s\cos (t\mathrm{/}2))(\sin (t\mathrm{/}2)\cos t,\sin (t\mathrm{/}2)\sin t,-\cos (t\mathrm{/}2))+\frac{s}{2}(-\sin t,\cos t,0)$, so $\mathrm{\parallel }{\mathrm{\Psi }}_t\times {\mathrm{\Psi }}_s{\mathrm{\parallel }}^2=(1+s\cos (t\mathrm{/}2){)}^2+s^2\mathrm{/}4$. Since $\mathrm{\mid }s\mathrm{\mid }\le 1\mathrm{/}2$ and $\mathrm{\mid }\cos (t\mathrm{/}2)\mathrm{\mid }\le 1$ by [L5], one has $1+s\cos (t\mathrm{/}2)\ge 1\mathrm{/}2$, so this norm squared is positive. For injectivity on either rectangle, equality of two images first gives the same polar angle $t$ modulo $2\pi$ because the radial coordinate is positive; the angle interval has length below $2\pi$, so the parameters have the same $t$. The radial coordinate together with the third coordinate $s\sin (t\mathrm{/}2)$ then recovers $s$, because ${\cos }^2(t\mathrm{/}2)+{\sin }^2(t\mathrm{/}2)=1$. Thus no interior parameter point shares its image with another point of the same rectangle, and both restrictions are regular.

The points of the first overlap that come from the interiors of both parameter regions have common parameters in $K_1^{\circ }=(5\pi \mathrm{/}6,7\pi \mathrm{/}6)\times (-1\mathrm{/}2,1\mathrm{/}2)$. On this open rectangle the restrictions are literally the same map with the same derivatives, so their oriented area vectors and induced normals agree by [F4]. At $(t,s)=(\pi ,0)$, step 3.1 gives ${\mathrm{\Psi }}_t\times {\mathrm{\Psi }}_s=(-1,0,0)$, so the common induced normal there is $-e_x$. No normal is asserted at an overlap point represented only by a boundary parameter, because [F4] defines the induced normal on the image of the parameter-region interior.

The points of the second overlap that come from both interiors are represented on $R_1$ by $K_2^{\circ }=(-\pi \mathrm{/}6,\pi \mathrm{/}6)\times (-1\mathrm{/}2,1\mathrm{/}2)$ and on $R_2$ by $h(K_2^{\circ })$, where $h(t,s)=(t+2\pi ,-s)$. Step 1.1 gives ${\mathrm{\Psi }}_2\circ h={\mathrm{\Psi }}_1$ there. Substituting $(t+2\pi ,-s)$ into the explicit formula of step 3.1 changes the sign of every term in ${\mathrm{\Psi }}_t\times {\mathrm{\Psi }}_s$, because $\sin ((t+2\pi )\mathrm{/}2)=-\sin (t\mathrm{/}2)$, $\cos ((t+2\pi )\mathrm{/}2)=-\cos (t\mathrm{/}2)$, $\sin (t+2\pi )=\sin t$, and $\cos (t+2\pi )=\cos t$. Thus the oriented area vectors, and hence the induced normals from [F4], are opposite at every such interior-overlap point. At $(t,s)=(0,0)$ on $R_1$, step 3.1 gives ${\mathrm{\Psi }}_t\times {\mathrm{\Psi }}_s=(0,0,-1)$, while the corresponding point $(2\pi ,0)$ on $R_2$ gives $(0,0,1)$.

Steps 4.1 and 4.2 exhibit, on the points where both induced normals are defined, one overlap component with matching normals and one with opposite normals. Thus this two-patch presentation carries both sign patterns at once.

Each overlap preimage is a closed rectangle of positive area, so the content-zero overlap condition in [F2] also fails. The example is therefore a two-patch presentation of the Möbius band, but not a compatible finite patch presentation for flux.