If $f:S^2 \rightarrow S^2$ is a Thurston mapping of degree $d$ (a finite ramified covering map with finite postcritical set $P_f$).
The Thurston pullback map $\sigma_f: \mathcal{T}_{S^2 \setminus P_f} \rightarrow \mathcal{T}_{S^2 \setminus P_f}$ on the Teichmüller space $\mathcal{T}_{S^2 \setminus P_f}$ is defined as follows (following the construction in Hubbard's Teichmüller Theory: Volume 2). Let $[\varphi] \in \mathcal{T}_{S^2 \setminus P_f}$. Then, a complex structure can be defined on $S^2 \setminus P_f$ by taking restrictions of $\varphi \circ f$ on open subsets not containing critical points, and at a critical point $z$, choosing a branch of $(\varphi \circ f)^{1/\text{deg}_z f}$. This gives a new complex structure such that $S^2$ is isomorphic to the Riemann sphere. Suppose $\varphi':S^2 \rightarrow \mathbb{P}^1$ is such an isomorphism. Define the Thurston pullback map by $$\sigma_f([\varphi])=[\varphi'].$$
To show this map is well-defined, suppose that $[\varphi], [\psi] \in \mathcal{T}_{S^2 \setminus P_f}$ with $[\varphi]=[\psi]$. Then, there exists a biholomorphism $\Phi:\mathbb{P}^1 \rightarrow \mathbb{P}^1$ such that $\psi$ is homotopic to $\Phi \circ \varphi$. How can we use this to show that $[\varphi']=[\psi']$?