$\DeclareMathOperator{\Gal}{Gal}$ Let $K$ be a field and $L/K$ and algebraic extension of $K$ $(L \subset \overline{K})$. Assume that for all $\sigma \in \Gal(\overline{K}/K)$, we have $\sigma\bigr|_L: L \to L \in \Gal(L/K)$. Show that $G(L/K) = \{\sigma\bigr|_L: \sigma \in G(\overline {K}/K) \}$
By hypothesis we have $ \{\sigma/_L: \sigma \in \Gal(\overline {K}/K) \} \subset \Gal(L/K)$. How can I prove the other side? That is, if I have $\tau \in G(L/K)$, how can I prove that there exists a $\sigma \in \Gal(\overline{K}/K)$ s.t $\sigma\bigr|_L = \tau$.
The only thing I know is: $\tau: L \to \overline{K}$ is injective.
I've seen this question before, It's very common to prove that in some books about Galois Theory, but the hypothesis that $L/K$ is algebraic is replaced by $L/K$ is finite. You can see a proof about that in the book ''Théorie de Galois'', by Josette Calais.
I could not use that proof to prove this exercise (Cause $L/K$ finite use another theorem), so can you help me?