Smoothing a continuous circle-valued map through an annular retraction

The map $F$ is continuous and well defined because the endpoint values and derivatives at $t=\pm \pi$ agree. Near $t=0$, $\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }$ has an absolute-value cusp, so $F$ fails to be smooth there.

Fix $0<\delta <1\mathrm{/}2$ and put $g_{\delta }(t)=\sqrt{{\sin }^2(t\mathrm{/}2)+{\delta }^2}-\delta$ and $H_{\delta }(e^{it})=(1-\delta )e^{i{g}_{\delta }(t)}$. Because ${\sin }^2(t\mathrm{/}2)=(1-\cos t)\mathrm{/}2$, the function $g_{\delta }$ is smooth and $2\pi$-periodic, so $H_{\delta }:S^1\to {\mathbb{R}}^2$ is smooth. Moreover, for $u=\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }$ one has $0\le u-g_{\delta }(t)\le \delta$, giving $\mathrm{\mid }{H}_{\delta }(e^{it})-F(e^{it})\mathrm{\mid }\le 2\delta$. Thus $H_{\delta }$ uniformly approximates the planar representative of $F$ as $\delta \downarrow 0$. Its image lies in the annulus $1\mathrm{/}2<\mathrm{\mid }z\mathrm{\mid }<3\mathrm{/}2$.

The radial retraction $R(z)=z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ is smooth on that annulus, and $\tilde{F}=R\circ H_{\delta }=e^{i{g}_{\delta }(t)}$ is a smooth circle-valued map. The formula $\mathcal{H}(e^{it},u)=\exp \text{\hspace{0.17em}⁣}\left(i\right((1-u)\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }+u{g}_{\delta }(t)\left)\right)$ for $0\le u\le 1$ is continuous and $2\pi$-periodic in $t$, so it defines a homotopy on $S^1$ from $F$ to $\tilde{F}$. This constructs the claimed smoothing and annular retraction directly.