field theory - smallest normal extension containing an infinite algebraic extension

Let $F/k$ be an algebraic extension. Let $S(F/k)$ be the set of all embeddings of F over k into algebraic closure $k^\mathrm{a}$. I'm trying to prove that the smallest normal extension of k containing F is

$E= \displaystyle\prod_{\sigma \in S(F/k)} \sigma F$ ($\Pi$ is used to denote the compositum of fields)

In the finite extension case, matter is simple as if $\tau$ is embedding of E over k, $\sigma \mapsto \tau\sigma$ is an injective mapping of $S(F/k)$ into $S(F/k)$. Since $S(F/k)$ is finite, the above mapping is bijective and $E= \displaystyle\prod_{\sigma \in S(F/k)} \sigma F=\prod{\tau\sigma F}$.

But in the infinite extension case when $S(F/k)$ might not be finite, surjective part must be added to prove that $\tau$ induces permutation on $S(F/k)$. I don't quite get an idea of how I should prove the surjectiveness.