Proof:

Using lemma (7.5) we begin by writing

$$h^\prime[\gamma] = [\gamma_0^\prime \ast \gamma \ast (\gamma_0^{\... ...mma_0^{\prime \prime})^{-1} \ast \gamma_0^{\prime \prime} \ast \gamma_0^{-1}].$$

By definition (7.2), the right hand side equals

$$[\gamma_0^\prime][\gamma_0^{\prime \prime}]^{-1}h^{\prime \prime}... ...{-1} =(\sigma \circ h^{\prime \prime} \circ \sigma^{-1})[\gamma] \eqno\square$$