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.
The easiest way to think about this is to not care about the induced map on $S(F/k)$ but instead just look at the field map. If $\tau:E\to k^\mathrm{a}$ is an embedding over $k$, then its image is contained in $E$, and so it gives an endomorphism of $E$ over $k$. Such an endomorphism is automatically surjective: if $E$ is an algebraic extension of a field $k$ and $\tau:E\to E$ is a homomorphism over $k$, then $\tau$ is surjective.
To prove this, consider and $a\in E$. The set $S$ of elements of $K$ which have the same minimal polynomial over $k$ as $a$ is finite, and $\tau$ maps $S$ to itself. Since $\tau$ is injective, this implies $\tau$ must restrict to a bijection from $S$ to itself. In particular, since $a\in S$, $a$ is in the image of $\tau$, and since $a$ was arbitrary, $\tau$ is surjective.