Show that $F_1\cap F_2/K$ is normal if $F_1/K$ and $F_2/K$ are normal. I have to show that $F_1\cap F_2/K$ is normal if $F_1/K$ and $F_2/K$ are normal. What I know is that \begin{align*}Aut_K(F_1)&\longrightarrow Hom_K(F_1\cap F_2,K^{alg})\\ \rho&\longmapsto \rho|_{F_1\cap F_2}\end{align*}
and
\begin{align*}Aut_K(F_2)&\longrightarrow Hom_K(F_1\cap F_2,K^{alg})\\ \rho&\longmapsto \rho|_{F_1\cap F_2}\end{align*}
are surjective. So let $\sigma\in Hom_K(F_1\cap F_2,K^{alg})$. 
Question) Since there is a $\rho\in Aut_K(F_1)$ s.t. $\rho|_{F_1\cap F_2}=\sigma $ can we conclude directely that $\rho|_{F_1\cap F_2}\in Aut_K(F_1\cap F_2)$ and thus that $\sigma \in Aut(K_1\cap K_2)$. Is it correct ?
 A: Let's recall what an algebraic normal extension $K \subset F$ is: it is an algebraic extension with the property that whenever an irreducible $P\in K[X]$ has a root in $F$, $P$ splits completely in $F[X]$. Equivalently, $F$ can be described as the union (possibly infinite) of all the roots of some family of polynomials in $K[X]$ lying in some algebraic closure of $K$. This family of course may be taken to consists only of irreducible polynomials.
Let now $F_1$ and $F_2$ irreducible. Let $\alpha \in F_1 \cap F_2$. Take $P_i$  an irreducible monic polynomial in $K[X]$ that has all the  roots in $F_i$ and moreover that has $\alpha$ as a root. Since $P_1$ and $P_2$ are irreducible in $K[X]$ and have a common root their gcd is not $1$. But their gcd is in $K[X]$ ( by Euclid) and it divides both $P_1$ and $P_2$. We conclude that $P_1 = P_2 = P$ and all the roots of $P$ are in $F_1 \cap F_2$. 
${\bf Added:}$ I see that the OP wanted a proof that involves morphisms of fields over $K$. 
We have $F_1$, $F_2$ contained in a common algebraic extension of $K$ so we may assume $F_1$, $F_2\subset \bar K^{a}$. The fact that some subextension $F/K$ of 
$\bar K^{a}/K$ is normal means: every map $F \to \bar K^{a}$ that is the identity on $K$ takes $F$ to itself. So now take $\phi$ mapping $F_1 \cap F_2$ to $\bar K^{a}$ and identity on $K$. Extend $\phi$ to an automorphism of $\bar K^{a}$. Since the $F_i$ are normal we have $\phi(F_i) \subset F_i$ and therefore $\phi(F_1 \cap F_2) \subset F_1 \cap F_2 $ ( in fact $=$). Therefore, $F_1 \cap F_2$ normal
The OP was on the right track, just needed to extend the automorphism all the way to something that contained $F_1 F_2$. 
A: What is your definition of a normal extension? (See here
 for options.)
Suppose $\sigma : F_1 \cap F_2 \to K^{\mathrm{alg}}$ is a $K$-embedding. You apparently know that there exists $\rho \in \mathrm{Aut}(F_1)$ such that $\rho|_{F_1 \cap F_2} = \sigma$. In particular, $\sigma(F_1 \cap F_2) = \rho(F_1 \cap F_2) \subseteq F_1$. Using a similar argument on $F_2$, you conclude $\sigma(F_1 \cap F_2) \subseteq F_2$, hence $\sigma(F_1 \cap F_2) \subseteq F_1 \cap F_2$. 
If $F_1$ or $F_2$ is finite over $K$, then so is $F_1 \cap F_2$, so you would be done using the dimension argument of $K$-vector spaces since $\sigma$ is injective and $K$-linear. 
If anyone can think of an argument which does not require using the second definition of normality (as displayed in my link above) in the case where $[F_1 \cap F_2:K] = \infty$, I would be pleased. In this case, the only argument I have is that $F_1/K$ is normal, thus $F_1$ is the union of all the finite normal extensions $F_1'/K$ where $F_1' \subseteq F_1$ (this is because splitting fields are normal and normal extensions contains the splitting fields of each of their elements, and I am using the second definition to prove this straightforwardly). 
Note that if you know the second definition, the proof of this statement becomes trivial. If a polynomial $p(X) \in (F_1 \cap F_2)[X]$ has a root in $F_1 \cap F_2$, then it has a root in $F_1$, hence all of its roots in $F_1$, and similarly for $F_2$. Thus $F_1 \cap F_2$ is normal. 
Hope that helps,
A: Let $P$ be a polynomial of $F_1\cap F_2$, suppose there exists $x\in F_1\cap F_2: P(x)=0$, since $F_i, i=1,2$ is normal, all the roots of $P$ are in $F_i$ so there are in $F_1\cap F_2$ which is henceforth normal. done.
