Suppose E \leq \bar{F} is an algebraic extension of of F and \alpha, \beta \in E are conjugate over F . Then the conjugation isomorphism \psi_{\alpha, \beta}: F(\alpha) \to F(\beta) can be extended to an isomorphism of E onto a subfield of \bar{F} .

So is it never always possible to extend it onto \bar{F} itself? Is \mathbb{C} its own algebraic closure? In other words, does \mathbb{C} = \bar{\mathbb{C}} ?