The once-around circle path is not thin-contractible

The map $f(x,y)=x^2+y^2$ has derivative $(2x,2y)$, which is nonzero on $f^{-1}(1)$; thus $1$ is a regular value and [F2] makes $S^1$ a smooth one-dimensional submanifold. Let $\tilde{\alpha }=xdy-ydx$ be the ambient one-form on ${\mathbb{R}}^2$, let $\iota :S^1\hookrightarrow {\mathbb{R}}^2$ be inclusion, and put $\alpha ={\iota }^{\ast }\tilde{\alpha }$, which is defined by [F8] and smooth by [F9]. The global tangent field $X=(-y,x){\mathrm{\mid }}_{S^1}$ spans each tangent line and satisfies $\alpha (X)=x^2+y^2=1$. By [F10], $d\alpha$ is a smooth two-form. By [F4], $d\alpha (X,X)=X(\alpha (X))-X(\alpha (X))-\alpha ([X,X])=0$; since $X$ spans each one-dimensional tangent space, this gives $d\alpha =0$ intrinsically on $S^1$. Along $\gamma$, $\dot{\gamma }(t)=2\pi X_{\gamma (t)}$, so [F6] gives ${\int }_{\gamma }\alpha ={\int }_0^{1}2\pi dt=2\pi$; the constant loop has integral zero.

Suppose a thin homotopy $H:[0,1{]}^2\to S^1$ joined $\gamma$ to the constant loop while fixing the basepoint at the path endpoints. Parametrize every curvilinear triangle in its finite subdivision by the specified smooth map from the standard two-simplex and compose that parametrization with $H$. By the smooth-up-to-boundary convention of Piecewise-smooth paths and thin homotopies, each composition has a smooth extension to a neighborhood of the standard simplex, so it is a smooth singular simplex by [F7]. Orient these finitely many simplices compatibly with the square and apply [F5] to each, then add the formulas. Each shared geometric edge occurs twice with opposite orientations. Choose one smooth reference parametrization $\sigma$ for such an edge; the two oriented face restrictions are $\sigma \circ \varphi$ and the reverse of $\sigma \circ \psi$, for increasing smooth reparametrizations $\varphi ,\psi :[0,1]\to [0,1]$. By [F6] and one-variable substitution, their oriented integrals are ${\int }_{\sigma }\alpha$ and $-{\int }_{\sigma }\alpha$, so they cancel. The remaining terms are exactly the oriented square boundary under $H$. Step 1.1 gives $d\alpha =0$, hence Stokes gives zero for the total boundary integral. The two path edges have signed integral $2\pi$ or $-2\pi$ because one is $\gamma$ and the other is constant; the endpoint edges are constant and contribute zero. Thus the boundary integral is $\pm 2\pi \mathrm{\ne }0$, a contradiction. Therefore $[\gamma ]$ is not the identity class.