= Solution
Put $w=g_T^{-1}(U_T+z)$ and define
$$
f_s(z)=g_{T-s}(w)-U_T.
$$
Then $f_0(z)=z$. Differentiating with the <Chordal Loewner equation> gives
$$
\partial_sf_s(z)=-\frac2{g_{T-s}(w)-U_{T-s}}=-\frac2{f_s(z)-(U_{T-s}-U_T)}.
$$
Uniqueness for this <ordinary differential equation> shows that $f_s=h_s$. At $s=T$,
$$
h_T(z)=w-U_T=g_T^{-1}(U_T+z)-U_T,
$$
which is the endpoint identity for the <Reverse Loewner flow>.
For $0<s<T$, the pathwise identity $h_s(z)=g_s^{-1}(U_s+z)-U_s$ is generally false. The left side is built from the reversed final driver segment $(U_{T-r}-U_T)_{0\leq r\leq s}$, whereas the right side is built from the initial segment $(U_r)_{0\leq r\leq s}$. By time reversal and symmetry of <Brownian motion> they have the same <probability distribution>, but they are not equal for the given Brownian path.
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