Why does surjective induced map imply surjective map?

Question

In the answer to this question: Showing a homomorphism of a field algebraic over $\mathbb{Q}$ to itself is an isomorphism. I am curious why the surjectivity of the induced map on the finite set implies that the map $\varphi$ is itself surjective. If you restrict the domain of a function, and that function is surjective, does this imply that the function itself is surjective? Or am I missing something?

Answer 1

This is just a statement of linear algebra. Since the map is $\mathbb Q$ linear, it is a $\mathbb Q$ linear endomorphism, of a finite dimensional vector space. Apply rank nullity, and deduce that the kernel must be $0$ dimensional, so the image has full rank.

In the link you show, the proof is quite nice as well. Let $\alpha \in F$ be arbitrary. We wish to find some $\alpha^{\prime}$ so that $\phi(\alpha^{\prime})=\alpha$. Let $m_{\alpha}$ be the minimal polynomial for $\alpha$, and $S$ all of its roots. Since $\phi:S \to S$ is a bijective map from roots of $m_\alpha$ for each $\alpha \in F$, it follows that there exists some $\alpha^{\prime}$ that is a root of $m_\alpha$ so that $\phi(\alpha^{\prime})=\alpha$.

Answer 2

The point here is that $\varphi$ maps every element $\alpha \in F$ to some root of the minimal polynomial of $\alpha$. So $\varphi$ permutes these roots. If you pick any element of $F$, it will be the image of some root of its minimal polynomial (otherwise the map induced on $S$ would not be surjective), so $\varphi$ is surjective.