Given that $g_t(z)$ is the solution for $\partial_tg_t(z) = \dfrac{2}{g_t(z) - \lambda(t)}$ with the initial value $g_o(z) = z$. How do I prove that, for fixed $t_0$, the function $h_{t,t_0}(z) = g_{t+t_0} \circ g_{t_0}^{-1}(z)$, satisfies $\partial _th_t(z) = \dfrac{2}{h_t(z) - \lambda(t_0 +t)}$
In better words, how would I prove that if $g_t(z)$ is the chordal Loewner map driven by $\lambda(t)$ then $h_{t,t_0}(z) = g_{t+t_0} \circ g_{t_0}^{-1}(z)$ is the chordal loewner map driven by $\lambda(t_0 +t)$
This is a specific part from a more general statement about chordal Loewner Equations in Basic properties of SLE (porposition 2.1). My main problem is how to differentiate the given composition.
Let's decompose the problem: $$ y=g_{t_0}^{-1}(z)~~\text{ solves }~~ z=g_{t_0}(y) $$ Then $$ h_{s,t_0}(z)=g_{t_0+s}(y). $$ So you identify solutions of the original problem by their (unique) value at time $t_0$ and then advance that solution to time $t_0+t$. The differential equation at time $t_0+t$ reads as $$ ∂_tg_{t_0+t}(y)=\frac{2}{g_{t_0+t}(y)+λ(t_0+t)} $$ Now replace back $h(z)$ for $g(y)$ to get $$ ∂_th_{t,t_0}(z)=\frac{2}{h_{t,t_0}(z)+λ(t_0+t)}. $$ So your transformation is just a re-parametrization of the original equation.