Continuing a square root once around the origin changes its sign

For $t\in [0,1]$, let $V_t:=e^{2\pi it}B(1,1\mathrm{/}2)$ and define $L_t(u):=2\pi it+\mathrm{Log}(e^{-2\pi it}u)$ and $R_t(u):=\exp (L_t(u)\mathrm{/}2)$. Because $e^{-2\pi it}{V}_t=B(1,1\mathrm{/}2)$ lies in the slit plane, [L1] makes each $R_t$ a holomorphic square-root branch on $V_t$, and $R_t(\gamma (t))=e^{\pi it}$.

Use the subdivision $t_k:=k\mathrm{/}16$ for $0\le k\le 16$. For $t\in [t_k,t_{k+1}]$, $\mathrm{\mid }\gamma (t)-\gamma (t_k)\mathrm{\mid }\le 2\sin (\pi \mathrm{/}16)<\frac{1}{2},$ so the whole subpath lies in $V_{t_k}$. At the joining point, the argument increment is $\pi \mathrm{/}8$, so the logarithm branches $L_{t_k}$ and $L_{t_{k+1}}$ agree there and hence define the same germ; therefore their square-root branches do as well. Thus the $R_{t_k}$ form an admissible continuation chain along $\gamma$. At $t=0$ the value at $1$ is $1$, while the terminal value is $e^{\pi i}=-1$. So the terminal germ is the negative of the initial square-root germ.