The fundamental-group label controls contractibility of the Whitney circle

If $F:D^2\to X$ fills $\gamma$, with the marked boundary point $z_{\ast }=1$ mapping to $p$, then $H(s,t)=F((1-t)e^{2\pi is}+t{z}_{\ast })$ is a based nullhomotopy: the segment stays in the convex disk, $H(s,0)=\gamma (s)$, $H(s,1)=p$, and $H(0,t)=H(1,t)=p$. Conversely a based nullhomotopy $H$ descends through the quotient $(s,t)\mapsto (1-t)e^{2\pi is}$, because its two side edges agree and its top edge is constant. The descended continuous map fills $\gamma$. Thus $\gamma$ bounds a continuous disk exactly when $[\gamma ]=1$.

For another $A$ arc $\alpha \mathrm{\prime }$, cancellation of $\bar{\alpha }\ast \alpha$ gives $[\alpha \mathrm{\prime }\ast \beta ]=[\alpha \mathrm{\prime }\ast \bar{\alpha }][\gamma ]$. The first factor is the image of a loop in $A$ based at $p$. For another $B$ arc $\beta \mathrm{\prime }$, the loop $\delta =\beta \mathrm{\prime }\ast \bar{\beta }$ is based at $q$, and $[\alpha \ast \beta \mathrm{\prime }]=[\alpha \ast \delta \ast \bar{\alpha }][\gamma ]$. Thus this change is a left multiplier in the image of the $B$ group transported from $q$ to $p$ along $\alpha$. All displayed products now lie in ${\pi }_1(X,p)$.

The continuous criterion needs no smoothing. For a smooth refinement, reparametrize each smooth arc by a smooth increasing interval map flat to all orders at both endpoints. Their concatenation $\tilde{\gamma }$ is a smooth based loop, since all one-sided derivatives vanish at both corners; interpolation of the interval parameters gives a based homotopy to $\gamma$. This reparametrized loop need not be an immersion. A continuous filling of $\tilde{\gamma }$ can be made radial-constant on an outer annulus, extended a little outside the disk, and smoothed by [F3] relative to a closed exterior annulus. Its restriction is a smooth disk map with that boundary. This optional refinement assumes ${\mathrm{A}\mathrm{C}}_{\omega }$; it does not assert the product corner collars or cleanliness required of a Whitney disk, which are supplied by the later geometric constructions.

For (iii), fix a base point $x_0\in X$ and a path $\lambda$ from $x_0$ to $p$, and for a double point $z$ choose a path $a_z$ in $A$ from $p$ to $z$ and a path $b_z$ in $B$ from $z$ to $p$; define the group label $g(z):=[\lambda \ast a_z\ast b_z\ast \bar{\lambda }]\in {\pi }_1(X,x_0)$. With the compatible choices $a_p,b_p$ constant, $a_q=\alpha$, $b_q=\beta$ this gives $g(p)=[\lambda \ast \bar{\lambda }]=1$ and $g(q)=[\lambda \ast \alpha \ast \beta \ast \bar{\lambda }]={\lambda }_{\ast }[\gamma ]$, so $g(q)=g(p)[\gamma ]$ after transporting back along $\lambda$; changing $a_z$ along a loop of $A$ or $b_z$ along a loop of $B$ multiplies the label by an element of the image of the corresponding sheet group, which is the path convention left open in the statement. Hence the two points carry equal labels exactly when $[\gamma ]=1$, which by step 1.1 is exactly the condition that the Whitney circle bounds a disk; When $X,A,B$ are oriented and the signs are opposite, [F5] defines $\epsilon (z)$, and $I(z):=\epsilon (z)g(z)$ satisfies $I(p)=-I(q)$ exactly when the labels are equal. The unsigned label equality and disk criterion do not require orientations. If $X$ is simply connected, then ${\pi }_1(X,p)=1$ by [F4] and $[\gamma ]=1$ for every choice of arcs, so every Whitney circle is null-homotopic.

Finally, the signed hypothesis alone is strictly weaker. For the declared smooth-torus supplier [F5] in this witness assume ${\mathrm{A}\mathrm{C}}_{\omega }$. In the torus $X_0=T^2={\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$ oriented as the product of its two circle factors let $A_0$ be the horizontal circle $\mathbb{R}\times \{0\}\mathrm{/}{\mathbb{Z}}^2$ and let $B_0$ be the graph of $f(x)=\frac{1}{4}\sin (2\pi x)$; orient both circles by increasing $x$. They are closed connected embedded circles, so they are complementary in dimension $2$, and $A_0\cap B_0=\{(0,0),(1\mathrm{/}2,0)\}$ with transverse crossings because $f$ vanishes exactly at $x=0$ and $x=1\mathrm{/}2$ in $[0,1)$ and $f^{\mathrm{\prime }}(0)=\pi \mathrm{/}2>0>f^{\mathrm{\prime }}(1\mathrm{/}2)$. In the frame $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ the isomorphism $T_p{A}_0\oplus T_p{B}_0\to T_p{X}_0$ has matrix with columns $(1,0)$ and $(1,f^{\mathrm{\prime }}(x))$, of determinant $f^{\mathrm{\prime }}(x)$, so by [F5] the intersection signs are $\epsilon (0,0)=+1$ and $\epsilon (1\mathrm{/}2,0)=-1$: the two points have opposite signs and the signed count vanishes. Let $\alpha (t)=(t\mathrm{/}2,0)$ and $\beta (t)=((1+t)\mathrm{/}2,f((1+t)\mathrm{/}2))$ for $t\in I$, using quotient coordinates; then $\gamma =\alpha \ast \beta$ is a Whitney circle for the pair. Its first coordinate traces $0\mapsto 1\mathrm{/}2$ along $\alpha$ and $1\mathrm{/}2\mapsto 1$ along $\beta$, so the projection $pr:T^2\to \mathbb{R}\mathrm{/}\mathbb{Z}$ satisfies $pr\circ \gamma ={\omega }_1$, whose class is nontrivial by [F5]. If $[\gamma ]$ were trivial, then by the induced homomorphism of [F4] the class $[pr\circ \gamma ]=p{r}_{\ast }[\gamma ]$ would be trivial as well, a contradiction; hence $[\gamma ]\mathrm{\ne }1$ and, by step 1.1, the circle bounds no disk, although its two points have opposite signs and vanishing signed count. Therefore the signed cancellation hypothesis alone does not make the Whitney circle contractible; the missing datum is exactly the label of step 2.2.