Let $K\subset F\subset E\subset C=\bar{K}$ be field extensions where $C$ is algebraic closure of $K$. Then there exists a bijection $\text{Hom}(E/K,C/K)\to \text{Hom}(F/K,C/K)\times \text{Hom}(E/F,C/F)$ where $\text{Hom}$ are sets of $K-$algebra homomorphisms.
I understood the proof. However there is a saddle point of I do not get why I naturally obtain a map as the following.
Given any $f\in \text{Hom}(E/K,C/K)$, I can extend this to an isomorphism $f'$ of $\text{Hom}(C/K,C/K)$. So given $f\in \text{Hom}(F/K,C/K),\ g\in \text{Hom}(E/F,C/F)$, I obtain $f'\in \text{Hom}(C/K,C/K), \ g'\in \text{Hom}(C/F,C/F)$. In order to demonstrate a bijection, I can defined $\text{Hom}(F/K,C/K)\times \text{Hom}(E/F,C/F)\to \text{Hom}(E/K,C/K)$ by $(f,g)\to (f'\circ g')$.
I also have another choice by defining $g'\circ f'$ why this choice is eliminated at the beginning. This is more or less post-fact check that this does not work.
Why do I naturally choose $(f,g)\to f'\circ g'$? Is it correct to think that to extend to the smaller ground field, I always go from top to bottom rather than from bottom to top.
Or was there a natural reason to choose this morphism to start with?