Question is to prove that every Galois group $\text{Gal}(F)$ is a profinite group.
Galois group of a field $F$ is Galois group of extension $F^s$ (maximal seperable extension) over $F$.
Profinite group is a closed subgroup of product of finite groups (discrete topology) with product topology.
We want to see $\text{Gal}(F)$ as subgroup of product of finite groups.
Consider finite galois extensions $K/F$, we have $F\subseteq K\subseteq F^s$. As $K/F$ is finite, Galois group is finite. Consider product of (topological) groups $P_F=\prod_{F\subset K} \rm{Gal}(K/F)$ and the map $$\rm{Gal}(F)\rightarrow\prod_{F\subseteq K} \rm{Gal}(K/F)$$ with $\sigma\mapsto (\sigma|_K)$. Idea is to prove that $\text{Gal}(F)$ is intersection of closed subsets and those closed subsets comes from inverse image of graph of some continuous functions.
Fix a galois extension $F\subseteq K\subseteq L$ consider $\pi_{L,K}:P_F\rightarrow \text{Gal}(L/F)\times \text{Gal}(K/F)$ with projection on first coordinate and restriction on another coordinate i.e., $(\sigma)\mapsto (\sigma,\sigma|_K)$ where $(\sigma)$ is tuple in $P_F$ and $\sigma$ is the component corresponding to extension $L$ and $\sigma|_K$ is the restriction of this map to $K$.
This map is continuous and let $\Gamma_{L/K}$ be the graph of this map. Then the instructor writes
$$\bigcap_{F\subseteq K\subseteq L}\pi_{L,K}^{-1}(\Gamma_{L/K})=\text{Gal}(F)$$
I am not able to fill this gap. Is it atleast true that what i understood is correct? Intersection is over all extensions where $K,L$ both varies?