Plane arc extension and rectangular neighborhoods

The normalized arc and its double lift. Identify the plane with $\mathbb{C}$ and its compactification with the sphere. The Möbius homeomorphism $M(z)=(z-A)\mathrm{/}(z-B)$ takes the endpoints to $0,\mathrm{\infty }$. Its value at the original plane-infinity is $1$, which is not on the normalized arc $b=M\circ a$. For $0<t<1$, $b(t)$ lies in ${\mathbb{C}}^{\ast }$. Lift its argument continuously on that interval and set $c(t)=\mathrm{\mid }b(t){\mathrm{\mid }}^{1\mathrm{/}2}\exp (\mathrm{i}\arg (b(t))\mathrm{/}2)$. Such an argument is obtained by continuing the elementary local argument on successive compact subintervals. Then $c(t{)}^2=b(t)$, and $c$ is injective since $b$ is. Extend $c$ at the endpoints by $c(0)=0,c(1)=\mathrm{\infty }$: convergence of its modulus proves continuity even if its angle has no endpoint limit. The two arcs $c$ and $-c$ have disjoint interiors; equality $c(t)=-c(u)$ would give $b(t)=b(u)$, hence $t=u$ and $c(t)=0$, impossible in the interior. Their union $J$ is a Jordan curve on the sphere. The involution $\tau (z)=-z$ fixes $0,\mathrm{\infty }$ and exchanges these two arcs.

Why the involution exchanges the complementary disks. By [F1], the two closed complementary regions of $J$ are disks. Parametrize $J$ by $c(t)$ on one semicircle and $-c(t)$ on the other, with equal $t$ at reflected circle parameters. Its restriction $\tau {\mathrm{\mid }}_J$ is therefore circle reflection and acts as $-1$ on $H_1(J;\mathbb{Z})$ (reverse the oriented circle cycle). In contrast $\tau$ is a sphere rotation homotopic to the identity through $z\mapsto e^{\mathrm{i}\pi u}z$, so its action on $H_2(S^2;\mathbb{Z})$ is $+1$. If it preserved one complementary disk $U$, it would preserve the other disk $V$. Give the sphere its two-disk CW structure using [F1]. The pair sequence gives an isomorphism $H_2(S^2)\to H_2(S^2,\bar{V})$ since $\bar{V}$ is contractible. CW excision identifies this relative group with $H_2(\bar{U},J)$, and its boundary map to $H_1(J)$ is an isomorphism since $\bar{U}$ is a disk. Naturality [F2] would then force the action of $\tau$ on $H_1(J)$ to be $+1$, a contradiction. Hence $\tau$ exchanges the two complementary disks.

An equivariant relative extension. Set $r(t)=t\mathrm{/}(1-t)$, with $r(0)=0,r(1)=\mathrm{\infty }$, and prescribe $f(c(t))=r(t)$ and $f(-c(t))=-r(t)$. This is a homeomorphism $J\to \mathbb{R}\cup \{\mathrm{\infty }\}$ commuting with $\tau$. Choose one source disk $U$ and extend $f$ from its boundary to the closed upper hemisphere by [F1]: take the stereographic pole in the other source disk and a target pole in the lower hemisphere, so both relevant regions are bounded Jordan disks in their plane charts. Call this extension $F_{+}$. On the other source disk define $F_{-}=\tau \circ F_{+}\circ \tau$. Step 2.1 ensures this definition has the right domain and maps it to the lower hemisphere. On $J$ it agrees with $F_{+}$ because $f\tau =\tau f$. Pasting the two maps and their inverses gives a sphere homeomorphism $F$ commuting with $\tau$ and fixing $0,\mathrm{\infty }$.

Descending and restoring the plane point. The quotient of the sphere by $\tau$ is the sphere through the map $p(z)=z^2$, with $p(\mathrm{\infty })=\mathrm{\infty }$. Its fibers are exactly $\{z,-z\}$, and compactness makes $p$ a quotient map. Therefore $F$ and $F^{-1}$ descend to inverse sphere homeomorphisms $g$ with $g(b(t))=r(t{)}^2$. The point $v=g(1)$ lies outside the positive real ray $[0,\mathrm{\infty }]$, because $1\mathrm{\notin }b([0,1])$. Move $v$ to $-1$ by a homeomorphism $L$ fixing that ray pointwise. Here is an explicit existence construction: the ray complement is the slit plane with polar angle in $(0,2\pi )$. Rotate the polar angle of $v$ within this interval to $\pi$, then change its radius along the negative real axis to reach $-1$. Approximate this compact path by a finite polygonal path inside the open slit plane. Choose $\delta >0$ less than one third of the distance from this compact polygonal path to the closed positive ray. Subdivide its finitely many segments so each displacement $w$ has length below $\delta \mathrm{/}4$. At the current path vertex $x_0$, use $\eta (x)=\max (0,1-\parallel x-x_0\parallel \mathrm{/}\delta )$ and the map $x\mapsto x+w\eta (x)$. It moves $x_0$ to the next vertex, is supported in the closed $\delta$-ball about $x_0$, and $\parallel w\parallel \mathrm{Lip}(\eta )<1\mathrm{/}4$. The finitely many supports form a compact subset of the ray complement. They are injective by this bound and surjective by [F3] applied to the contraction equation $x=y-w\eta (x)$, Their inverses are Lipschitz with constant at most $1\mathrm{/}(1-\parallel w\parallel \mathrm{Lip}(\eta ))$, by the same lower distance bound. Thus finite small translations move the point along the path and fix its complement. This constructs $L$ supported away from the ray. Define the Möbius homeomorphism $T(z)=z\mathrm{/}(1+z)$, with $T(-1)=\mathrm{\infty }$ and $T(\mathrm{\infty })=1$. Now $TLgM$ fixes the original sphere-infinity, since its successive images are $1,v,-1,\mathrm{\infty }$, hence restricts to a plane homeomorphism, and takes $a(t)$ to $r(t{)}^2\mathrm{/}(1+r(t{)}^2)$.

The prescribed parameter and neighborhoods. The increasing homeomorphism $\lambda (t)=r(t{)}^2\mathrm{/}(1+r(t{)}^2)$ of $[0,1]$ has fixed endpoints. Extend ${\lambda }^{-1}$ to an increasing homeomorphism $\mathrm{\Lambda }$ of $\mathbb{R}$ equal to the identity outside $[0,1]$. Postcompose step 4.1 with $(x,y)\mapsto (\mathrm{\Lambda }(x),y)$ to obtain $G(a(t))=(t,0)$. Transport straight rectangular and endpoint-sector neighborhoods by $G^{-1}$. This proves the conclusion, with AC used precisely in the relative Jordan–Schönflies extensions of steps 2.1 and 3.1. No collar was inferred merely from connectivity of an arc complement.