Suppose $F\subseteq L$ is any field extension, $f(x) \in F[x]$ and $\beta_1,\beta_2,....\beta_r\in L$ are distinct roots of $f(x)$. Prove
a)If $\sigma$ is an automorphism of L that leaves F fixed pointwise, then $\sigma|_{\{\beta_1,...,\beta_r\}}$ is a permutation of $\{\beta_1,....,\beta_r\}$
b)$\sigma \mapsto \sigma|_{\{\beta_1,.....,\beta_r\}} $ is a homomorphism of $Aut_F(L)$ into the group of permutation $Sym(\{\beta_1,...,\beta_r\})$
c) If L is a splitting field of $f(x)$, $L=K(\beta_1,...., \beta_r)$, then the homomorphism $\sigma \mapsto \sigma |_{\{\beta_1,....\beta_r\}}$ is injective. Please help me understand.
My attempt: for a)Assume f(x) is irreducible in F[x]. $\sigma $ is an automorphism of L and $\beta_i$ is a root of $f(x)$ $F\subset F(\beta_1)\subset F(\beta_1,\beta_2)\subset .....\subset F(\beta_1,...,\beta_r)=L$. Hence $\sigma (\beta_i)\mapsto \beta_j$ and they are all distinct. That is $\sigma |_{\{\beta_1,...\beta_r\}}$ is a permutation of the roots.
For b, Let $\tau : Aut_F(L)\rightarrow Sym(\{\beta_1,...,\beta_r\})$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sigma \mapsto \sigma|_{\{\beta_1,...,\beta_r\}}$
That is $\tau (\sigma)=\sigma|_{\{\beta_1,...\beta_r\}}$ but I do not know how to do further.
HINT: For a., prove that $\sigma(\beta_i)$ must be a root of $f(x)$ for all $i$, thus $\sigma(\beta_i)=\beta_j$, since $\sigma$ is bijective $\cdots$. This (almost) immediately implies b.
For c. Prove that any automorphism of $F(\beta_1,...,\beta_r)$ that fixes $F$ is completely determined by its values on $\beta_1,...\beta_r$.