Let $G$ and $L$ be topological groups, and $G$ acts on $L$ via the map $f:G×L→L$. I want to prove $f$ is continuous. From definition,$f$ is continuous if only if for arbitrary open subset $U$ of $L$, $f^{-1}(U)$ is open in $G×L$. But my book goes like this,
For arbitrary point $a∈L$, let fundamental system of neighborhood be $V_a$. If I could find some open subgroup $H$ of $G$ such that $f(H,a)⊆V_a$, we can say $f$ is continuous.
But why? This argument just showed $H\times V_a⊆f^{-1}(V_a)$, what about other direction? And I wonder this kind of argument is naturally done.
c.f
↓This is my book, in the above notation, $G＝Gal(\overline{\Bbb{Q}_p}/\Bbb{Q}_p)$, $L＝\Bbb{C}_p$.
The book uses something more than what you write.
It shows the following:
For any $a \in L$ and any $g \in G$, and any $V$ from an open basis of $\sigma(a)$, there exists an open subgroup $H \subset G$ and an open neighborhood $U$ of $a$ such that $f(gH, U) \subseteq V$. Note that $gH \subseteq G$ is open, because $G$ is a topological group.
This suffices to conclude that $f$ is continuous, because for an arbitrary open set $V \subset L$ and any element $(g, a) \in f^{-1}(V)$, there is a fundamental open set $W$ such that $f(g, a) \in W \subseteq V$ and the above result gives us $H \subset G$ and $U\subset L$, both open, with $f(gH, U) \subseteq W \subseteq V$.
Therefore $gH \times U$ is an open neighborhood of $(g, a)$ in $f^{-1}(V)$. Since $(g, a) \in f^{-1}(V)$ is arbitrary, we see that $f^{-1}(V)$ is open.