Simply Connected Plane Domains: the Grand Equivalence

By [L1], the image of the homotopy square lies in $\mathrm{\Omega }$. For each point $x$ of $[0,1{]}^2$, openness of $\mathrm{\Omega }$ gives an open disc $D_x\subseteq \mathrm{\Omega }$ centered at $H(x)$, and the sets $H^{-1}(D_x)$ form an open cover of the compact square $[0,1{]}^2$. By [L2], there is $\delta >0$ such that every subset of the square of diameter less than $\delta$ lies in some $H^{-1}(D_x)$. Choose $N$ with $\sqrt{2}\mathrm{/}N<\delta$.

For $1\le j,k\le N$, write $Q_{jk}=[\frac{j-1}{N},\frac{j}{N}]\times [\frac{k-1}{N},\frac{k}{N}]\mathrm{.}$ Each cell has diameter $\sqrt{2}\mathrm{/}N<\delta$, so step 1.1 places $H[Q_{jk}]$ inside an open disc $D_{jk}\subseteq \mathrm{\Omega }$. Put $a_{jk}=H(\frac{j-1}{N},\frac{k-1}{N}),b_{jk}=H(\frac{j}{N},\frac{k-1}{N}),c_{jk}=H(\frac{j}{N},\frac{k}{N}),d_{jk}=H(\frac{j-1}{N},\frac{k}{N})\mathrm{.}$ Since every disc is convex, the straight segments from $a_{jk}$ to $b_{jk}$, from $b_{jk}$ to $c_{jk}$, from $c_{jk}$ to $d_{jk}$, and from $d_{jk}$ to $a_{jk}$ all lie in $D_{jk}$. Let $P_{jk}$ be the closed polygonal contour obtained by traversing those four segments in that order.

Because $D_{jk}$ is star-shaped, [L5] gives ${\int }_{P_{jk}}f(z)dz=0.$ Also [L3] gives a primitive $F_{jk}$ of $f$ on $D_{jk}$.

Summing the zero integrals from step 3.1 over all cells, every interior polygon edge appears once in each orientation, so [L6] and [L7] cancel all interior contributions. The surviving outer boundary is the bottom polygonal path $P_0$ built from the straight segments joining ${\gamma }_0((j-1)\mathrm{/}N)$ to ${\gamma }_0(j\mathrm{/}N)$, the top polygonal path $P_1$ built in the forward direction from the straight segments joining ${\gamma }_1((j-1)\mathrm{/}N)$ to ${\gamma }_1(j\mathrm{/}N)$ but occurring in the outer boundary with reverse orientation, and the two side edges. By [L1] both side edges are constant, and for a constant path $s$ one has $s\ast s=s$, so [L6] gives ${\int }_{s}fdz={\int }_{s\ast s}fdz=2{\int }_{s}fdz$, hence ${\int }_{s}fdz=0$. Therefore ${\int }_{P_0}f(z)dz={\int }_{P_1}f(z)dz\mathrm{.}$

For each $1\le j\le N$, the bottom subpath ${\gamma }_0{\mathrm{\mid }}_{[(j-1)\mathrm{/}N,j\mathrm{/}N]}$ and the chord segment from ${\gamma }_0((j-1)\mathrm{/}N)$ to ${\gamma }_0(j\mathrm{/}N)$ both lie in $D_{j1}$. Since $F_{j1}$ is a primitive of $f$ on $D_{j1}$, [L4] gives the same endpoint increment for both, so their integrals are equal. Summing over $j$ and using [L6] yields ${\int }_{{\gamma }_0}f(z)dz={\int }_{P_0}f(z)dz\mathrm{.}$ The same argument with the top-row discs $D_{jN}$ gives ${\int }_{{\gamma }_1}f(z)dz={\int }_{P_1}f(z)dz\mathrm{.}$

Combining steps 4.1 and 4.2 gives ${\int }_{{\gamma }_0}f(z)dz={\int }_{{\gamma }_1}f(z)dz,$ as required.

The Given and [L1] place $\gamma$ and the constant loop $c$ under the hypotheses of [L2]. Therefore ${\int }_{\gamma }f(z)dz={\int }_{c}f(z)dz\mathrm{.}$

Because $c$ is constant, one has $f(c(t))=f(c(0))$ for every $t$. Since [L4] makes $c$ a closed contour, [L3] gives ${\int }_{c}f(z)dz={\int }_{c}f(c(0))dz=f(c(0))(c(1)-c(0))=0.$ Combining this with step 1.1 proves ${\int }_{\gamma }f(z)dz=0$.

Because $\gamma (t)\mathrm{\ne }0$ for every $t$, the normalized map $\alpha (t)=\gamma (t)\mathrm{/}\mathrm{\mid }\gamma (t)\mathrm{\mid }$ is a continuous loop in the unit circle based at $1$. Using the homeomorphism of [L2], regard the same loop as a based loop $\beta :[0,1]\to \mathbb{R}\mathrm{/}\mathbb{Z}$ at $[0]$. Let $\tilde{\beta }:[0,1]\to \mathbb{R}$ be its lift with $\tilde{\beta }(0)=0$, and define $\theta (t)=2\pi \tilde{\beta }(t)$.

By the definition of $h$ in [L2], the lift from step 1.1 supplies a continuous argument. [step 1.1, L1, L2, algebra] $\alpha (t)=\cos \theta (t)+i\sin \theta (t)=e^{i\theta (t)}\mathrm{.}$ Hence $\gamma (t)=\mathrm{\mid }\gamma (t)\mathrm{\mid }{e}^{i\theta (t)},$ so $\theta$ is a continuous argument of $\gamma$ about $0$. Therefore [L1] gives $n(\gamma ,0)=\frac{\theta (1)-\theta (0)}{2\pi }=\tilde{\beta }(1)\mathrm{.}$

Since $\tilde{\beta }(1)$ is exactly the degree of $\beta$ by [L3], step 2.1 shows $n(\gamma ,0)=\mathrm{deg}(\beta )$, which is the asserted degree of the normalized circle loop of $\gamma$. Fact [L4] records that this is the same integer that classifies the loop class in the usual ${\pi }_1(S^1)\cong \mathbb{Z}$ convention.

Let $\mathrm{\Gamma }={\sum }_{j<r}{m}_j{\gamma }_j$ be a cycle with trace in $\mathrm{\Omega }$, and let $f$ be holomorphic on $\mathrm{\Omega }$. Choose a basepoint $z_0\in \mathrm{\Omega }$. Let $E:=\{{\gamma }_j(a_j):j<r,m_j\mathrm{\ne }0\}\cup \{{\gamma }_j(b_j):j<r,m_j\mathrm{\ne }0\}\mathrm{.}$ For each $q\in E$, [L5] gives a polygonal path ${\lambda }_q$ in $\mathrm{\Omega }$ from $z_0$ to $q$; by [L6] each ${\lambda }_q$ is a rectifiable contour.

Fix $j<r$ with $m_j\mathrm{\ne }0$, and write $a_j={\gamma }_j(a_j)$ and $b_j={\gamma }_j(b_j)$. The contour ${\delta }_j:={\lambda }_{a_j}\ast {\gamma }_j\ast {\lambda }_{b_j}^{-}$ is a based loop at $z_0$. By the triviality hypothesis, its loop class is the identity, so [L1] makes ${\delta }_j$ path-homotopic relative to the endpoints to the constant loop at $z_0$. Applying [L2] and then [L6] gives $0={\int }_{{\delta }_j}f(z)dz={\int }_{{\lambda }_{a_j}}f(z)dz+{\int }_{{\gamma }_j}f(z)dz-{\int }_{{\lambda }_{b_j}}f(z)dz,$ hence ${\int }_{{\gamma }_j}f(z)dz={\int }_{{\lambda }_{b_j}}f(z)dz-{\int }_{{\lambda }_{a_j}}f(z)dz\mathrm{.}$

By [L4] and step 1.2, ${\int }_{\mathrm{\Gamma }}f(z)dz={\sum }_{\begin{array}{c}j<r\\ m_j\mathrm{\ne }0\end{array}}{m}_j{\int }_{{\gamma }_j}f(z)dz={\sum }_{\begin{array}{c}j<r\\ m_j\mathrm{\ne }0\end{array}}{m}_j{\int }_{{\lambda }_{b_j}}f(z)dz-{\sum }_{\begin{array}{c}j<r\\ m_j\mathrm{\ne }0\end{array}}{m}_j{\int }_{{\lambda }_{a_j}}f(z)dz\mathrm{.}$ Grouping the two finite sums by endpoint $q\in E$ and using the boundary formula from [L4], this becomes ${\int }_{\mathrm{\Gamma }}f(z)dz={\sum }_{q\in E}\mathrm{\partial }\mathrm{\Gamma }(q){\int }_{{\lambda }_q}f(z)dz=0,$ because $\mathrm{\Gamma }$ is a cycle. Thus ${\int }_{\mathrm{\Gamma }}f(z)dz=0$ for every holomorphic $f$ and every cycle $\mathrm{\Gamma }$ in $\mathrm{\Omega }$.

The criterion in [L3] now shows that $\mathrm{\Omega }$ is homologically simply connected.

Since ${\mathrm{\Gamma }}^{\ast }\subseteq \mathrm{\Omega }$, the index $n(\mathrm{\Gamma },p)$ is defined for every $p\in \mathbb{C}\setminus \mathrm{\Omega }$. By [L1], the function $p\mapsto n(\mathrm{\Gamma },p)$ is locally constant on $\mathbb{C}\setminus {\mathrm{\Gamma }}^{\ast }$, and there is $R>0$ with $n(\mathrm{\Gamma },p)=0$ whenever $\mathrm{\mid }p\mathrm{\mid }>R$. Thus the subset [given, L1, construct] $E=\{\mathrm{\infty }\}\cup \{p\in \mathbb{C}\setminus \mathrm{\Omega }:n(\mathrm{\Gamma },p)=0\}$ contains $\mathrm{\infty }$ together with a punctured neighborhood of $\mathrm{\infty }$ in the sphere.

The set $E$ is open in $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ by the local constancy from [L1], and its complement in $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is open for the same reason. Since $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is connected and $E$ is nonempty by step 1.1, it follows that [step 1.1, L1, algebra] $E=\widehat{\mathbb{C}}\setminus \mathrm{\Omega }\mathrm{.}$ Therefore $n(\mathrm{\Gamma },p)=0$ for every $p\in \mathbb{C}\setminus \mathrm{\Omega }$.

By [L2], the vanishing from step 2.1 is exactly the statement that $\mathrm{\Gamma }$ is null-homologous in $\mathrm{\Omega }$.

Suppose, toward a contradiction, that $F:=\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is disconnected. Then $F=A\bigsqcup B$ for disjoint nonempty closed subsets $A,B\subseteq F$. Since $\mathrm{\infty }\in F$, relabel so that $\mathrm{\infty }\in B$. Because the Riemann sphere is a metric space, the disjoint closed sets $A$ and $B$ have disjoint open neighbourhoods $U$ and $V$ in $\widehat{\mathbb{C}}$ with $A\subseteq U$ and $B\subseteq V$. The inclusion $\mathrm{\infty }\in V$ forces $U\subseteq \mathbb{C}$, so $A$ is compact in $\mathbb{C}$ and $U\cap (\mathbb{C}\setminus \mathrm{\Omega })=U\cap F=A$. Hence $U\setminus A\subseteq \mathrm{\Omega }$.

Apply [L2] to the compact set $A\subseteq U$, viewing $\mathbb{C}$ as ${\mathbb{R}}^2$. It gives a finite union $J$ of closed grid rectangles with $A\subseteq \mathrm{int}J\subseteq J\subseteq U$. Give each rectangle boundary its positive orientation, sum those boundary chains, and cancel every interior edge with its opposite by [L3]. Let $\mathrm{\Gamma }$ be the remaining chain. Then $\mathrm{\Gamma }$ is a cycle, and its trace is the frontier of $J$, so ${\mathrm{\Gamma }}^{\ast }\subseteq U\setminus A\subseteq \mathrm{\Omega }$.

Let $p\in \mathrm{int}J$ lie on no grid line. Exactly one grid cell $Q$ of $J$ contains $p$. Along the positively oriented boundary of $Q$, the continuous argument of $\zeta -p$ increases by $2\pi$, while along the boundary of every other grid cell it has increment $0$ because $p$ lies outside that cell. Therefore [L4] gives winding number $1$ for ${\mathrm{\partial }}^{+}Q$ about $p$ and $0$ for every other cell boundary, and additivity from [L3] yields $n(\mathrm{\Gamma },p)=1$.

Fix $a\in A$. Because $A\subseteq \mathrm{int}J$, choose a disc $D(a,r)\subseteq \mathrm{int}J$ and then choose $p\in D(a,r)$ on no grid line. The disc misses ${\mathrm{\Gamma }}^{\ast }$, so local constancy from [L5] and step 3.1 give $n(\mathrm{\Gamma },a)=n(\mathrm{\Gamma },p)=1$.

The point $a$ lies in $\mathbb{C}\setminus \mathrm{\Omega }$, while step 2.1 gives ${\mathrm{\Gamma }}^{\ast }\subseteq \mathrm{\Omega }$. Step 4.1 yields $n(\mathrm{\Gamma },a)=1\mathrm{\ne }0$, so [L1] says that $\mathrm{\Gamma }$ is not null-homologous in $\mathrm{\Omega }$, contradicting the homological simple connectivity of $\mathrm{\Omega }$. Therefore the assumption of step 1.1 was false, and $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is connected.

Assume condition 1. Then [L1] gives condition 2 immediately.

Assume condition 2. Fix $p\in \mathbb{C}\setminus \mathrm{\Omega }$ and define [L3, given, choose, construct] $u_p(z)=\log \mathrm{\mid }z-p\mathrm{\mid }\mathrm{.}$ Direct differentiation gives $\frac{\mathrm{\partial }{u}_p}{\mathrm{\partial }x}=\frac{x-\mathrm{Re}p}{\mathrm{\mid }z-p{\mathrm{\mid }}^2},\frac{\mathrm{\partial }{u}_p}{\mathrm{\partial }y}=\frac{y-\mathrm{Im}p}{\mathrm{\mid }z-p{\mathrm{\mid }}^2},$ and then $\frac{{\mathrm{\partial }}^2{u}_p}{\mathrm{\partial }{x}^2}+\frac{{\mathrm{\partial }}^2{u}_p}{\mathrm{\partial }{y}^2}=0$ on $\mathrm{\Omega }$, because $p\mathrm{\notin }\mathrm{\Omega }$. So $u_p$ is harmonic on $\mathrm{\Omega }$. By condition 2 and [L3], choose a harmonic conjugate $v_p$ on $\mathrm{\Omega }$ and put $F_p=u_p+i{v}_p,$ which is holomorphic on $\mathrm{\Omega }$.

The function [step 1.2, L5, L6, algebra] $G_p(z)=\frac{\exp (F_p(z))}{z-p}$ is holomorphic on $\mathrm{\Omega }$ by [L5]. Its modulus is $\mathrm{\mid }{G}_p(z)\mathrm{\mid }=\frac{\mathrm{\mid }\exp (F_p(z))\mathrm{\mid }}{\mathrm{\mid }z-p\mathrm{\mid }}=\frac{e^{\mathrm{Re}{F}_p(z)}}{\mathrm{\mid }z-p\mathrm{\mid }}=\frac{e^{u_p(z)}}{\mathrm{\mid }z-p\mathrm{\mid }}=1,$ so $G_p(\mathrm{\Omega })$ lies on the unit circle. By [L6], $G_p$ cannot be nonconstant, hence it is constant: $G_p(z)\equiv c_p,\mathrm{\mid }{c}_p\mathrm{\mid }=1.$ Therefore $L_p:=F_p-\log c_p$ satisfies $\exp (L_p(z))=z-p$ on $\mathrm{\Omega }$.

By [L4], each $L_p$ from step 2.1 is a primitive of $1\mathrm{/}(z-p)$ on $\mathrm{\Omega }$. Since $p\in \mathbb{C}\setminus \mathrm{\Omega }$ was arbitrary, condition 4 of [L2] holds for $\mathrm{\Omega }$, so [L2] gives condition 1. Together with step 1.1, this proves the equivalence.

By [L1], $\mathrm{\Omega }$ is a complex domain. If $\mathrm{\Omega }=\mathbb{C}$, then the first alternative holds and there is nothing more to prove.

If $\mathrm{\Omega }\mathrm{\ne }\mathbb{C}$, then $\mathrm{\Omega }⊊\mathbb{C}$ is a proper homologically simply connected complex domain, so [L2] applies and gives a biholomorphic map $\mathrm{\Omega }\to \mathbb{D}$. This is exactly the second alternative.

Steps 1.1 and 2.1 prove the dichotomy.

Both $\mathbb{C}\cong {\mathbb{R}}^2$ and the unit disc $\mathbb{D}$ are convex subsets of ${\mathbb{R}}^2$, so [L1] makes $E$ contractible. In particular, the identity map ${\mathrm{id}}_E$ is homotopic to a constant map $c_e$ for some $e\in E$.

Postcompose the homotopy from step 1.1 by $h^{-1}$ and precompose it by $h$. By [L2], this yields a homotopy from [step 1.1, L2, algebra] $h^{-1}\circ {\mathrm{id}}_E\circ h={\mathrm{id}}_{\mathrm{\Omega }}$ to the constant map at $h^{-1}(e)$. Thus ${\mathrm{id}}_{\mathrm{\Omega }}$ is nullhomotopic.

Let $f:\mathrm{\Omega }\to Y$ be any continuous map into any topological space $Y$. Postcomposing the nullhomotopy from step 2.1 by $f$ and using [L2] again shows that [step 2.1, L2, L3] $f=f\circ {\mathrm{id}}_{\mathrm{\Omega }}$ is homotopic to the constant map at $f(h^{-1}(e))$. Hence every continuous map out of $\mathrm{\Omega }$ is nullhomotopic, so [L3] makes $\mathrm{\Omega }$ contractible. ∎

By [L1], there are a point $x_1\in X$ and a homotopy $H:X\times I\to X$ from ${\mathrm{id}}_X$ to the constant map $c_{x_1}$. Evaluating at the chosen basepoint gives a path $\lambda (t)=H(x_0,t)$ from $x_0$ to $x_1$.

Let $[\alpha ]\in {\pi }_1(X,x_0)$. Define $K:I\times I\to X$ by $K(s,t)=\{\begin{array}{ll}H(x_0,3st),& 0\le s\le \frac{1}{3},\\ H(\alpha (3s-1),t),& \frac{1}{3}\le s\le \frac{2}{3},\\ H(x_0,(3-3s)t),& \frac{2}{3}\le s\le 1.\end{array}$ The three pieces are continuous and agree on the seams $s=\frac{1}{3},\frac{2}{3}$ because $\alpha (0)=x_0=\alpha (1)$, so $K$ is continuous. Also $K(0,t)=x_0=K(1,t)$ for every $t$, so $K$ is a path homotopy relative to the endpoints between loops at $x_0$. At $t=0$ it is $c_{x_0}\ast \alpha \ast c_{x_0}$, and at $t=1$ it is $\lambda \ast c_{x_1}\ast \bar{\lambda }$. Hence $[c_{x_0}\ast \alpha \ast c_{x_0}]=[\lambda \ast c_{x_1}\ast \bar{\lambda }]\mathrm{.}$ Since $[c_{x_0}]$ is the identity by [L2], this gives $[\alpha ]=[\lambda \ast c_{x_1}\ast \bar{\lambda }]$.

Define $J:I\times I\to X$ by $J(s,u)=\{\begin{array}{ll}\lambda (3s(1-u)),& 0\le s\le \frac{1}{3},\\ \lambda (1-u),& \frac{1}{3}\le s\le \frac{2}{3},\\ \lambda ((3-3s)(1-u)),& \frac{2}{3}\le s\le 1.\end{array}$ The three pieces are continuous and agree on the seams, so $J$ is continuous. Also $J(0,u)=x_0=J(1,u)$ for every $u$, while $J(-,0)=\lambda \ast c_{x_1}\ast \bar{\lambda }$ and $J(-,1)=c_{x_0}$. Thus $[\lambda \ast c_{x_1}\ast \bar{\lambda }]=[c_{x_0}]$ by [L2], and step 2.1 yields $[\alpha ]=[c_{x_0}]$.

Every loop class at $x_0$ is therefore the identity, so ${\pi }_1(X,x_0)$ is trivial.

By [L1], conditions 1 and 2 are equivalent. By [L2], conditions 2, 4, and 6 are equivalent. By [L4], conditions 2 and 8 are equivalent. By [L9], conditions 4 and 5 are equivalent.

Condition 2 implies condition 7 by [L3]. Conversely, assume condition 7. Fix $p\in \mathbb{C}\setminus \mathrm{\Omega }$ and a closed rectifiable contour $\gamma :[a,b]\to \mathrm{\Omega }$. Applying condition 7 repeatedly to the nowhere-zero holomorphic function $z\mapsto z-p$ produces, for every $m\ge 1$, a nowhere-zero holomorphic function $g_m$ on $\mathrm{\Omega }$ with $g_m^{2^m}=z-p$. By [L10], the closed contour $g_m\circ \gamma$ admits a continuous logarithm ${\lambda }_m$; write ${\theta }_m=\mathrm{Im}{\lambda }_m$. Then $2^m{\lambda }_m$ is a continuous logarithm of $\gamma -p$ along $\gamma$, so [L11] gives $n(\gamma ,p)=2^m\frac{{\theta }_m(b)-{\theta }_m(a)}{2\pi }\mathrm{.}$ Applying [L11] again to $g_m\circ \gamma$ shows that $({\theta }_m(b)-{\theta }_m(a))\mathrm{/}(2\pi )$ is an integer. Therefore $n(\gamma ,p)\in 2^m\mathbb{Z}(m\ge 1)\mathrm{.}$ The only integer divisible by every power of $2$ is $0$, so $n(\gamma ,p)=0$. By [L11], this is equivalent to ${\int }_{\gamma }\frac{dz}{z-p}=0.$ Since $p$ and $\gamma$ were arbitrary and $z\mapsto 1\mathrm{/}(z-p)$ is continuous on $\mathrm{\Omega }$, [L9] makes $1\mathrm{/}(z-p)$ admit a primitive on $\mathrm{\Omega }$. Thus condition 2 holds by [L2], and conditions 2 and 7 are equivalent.

Assume condition 2. By [L5], condition 9 follows. Any conformal equivalence is in particular a homeomorphism, so condition 9 implies condition 10. Then [L6] gives condition 11, [L7] gives condition 3, and [L8] returns to condition 2. Thus [L5, L6, L7, L8] $2\Rightarrow 9\Rightarrow 10\Rightarrow 11\Rightarrow 3\Rightarrow 2.$

Steps 1.1, 1.2, and 1.3 connect every listed clause to condition 2, so all eleven conditions are equivalent.

If $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is connected, then [L2] gives homological simple connectivity, and [L1] identifies that with simple connectivity under the current Axiom-of-Choice hypothesis.

If $\mathrm{\Omega }$ is simply connected, then [L1] places it under the grand-equivalent conditions, so it is homologically simply connected; [L2] then gives connected spherical complement.

Steps 1.1 and 1.2 prove both directions.

Let $C=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }=1\}\subseteq {\mathbb{C}}^{\times }$ and let $r:{\mathbb{C}}^{\times }\to C$ be radial normalization, $r(z)=z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$. Specializing [L2] to $n=2$ and applying [L3], the induced map $r_{\ast }:{\pi }_1({\mathbb{C}}^{\times },1)⟶{\pi }_1(C,1)$ is an isomorphism. For the given loop $\gamma$, its image under $r_{\ast }$ is the class of the normalized circle loop $\alpha (t)=\gamma (t)\mathrm{/}\mathrm{\mid }\gamma (t)\mathrm{\mid }$.

Fact [L4] identifies the class of $\alpha$ in ${\pi }_1(C,1)$ with the integer $\mathrm{deg}(\alpha )$. Hence $W([\gamma ])=\mathrm{deg}(r\circ \gamma )$ is exactly the composite of the isomorphism $r_{\ast }$ from step 1.1 with the standard identification ${\pi }_1(C,1)\cong \mathbb{Z}$, and is therefore the standard isomorphism ${\pi }_1({\mathbb{C}}^{\times },1)\cong (\mathbb{Z},+)\mathrm{.}$ If $\gamma$ is rectifiable, [L1] gives $W([\gamma ])=\mathrm{deg}(\alpha )=n(\gamma ,0)$.