Let $K$ be a field. Let $G_K := Gal(\overline{K}/K)$ be absolute Galois group of $K$.
Let $M$ be a $G_K$-module, that is, $G_K$ (with the Krull topology) acts continuously on $M$ with discrete topology.
Fix $m\in M$. Then $f: G_K \to M, \sigma \to \sigma m-m$ is 1-cocycle, that is, it satisfies $f(\sigma \tau)=f(\sigma)+\sigma f(\tau)$.
Then why is $f$ continuous?
My book (Silverman's The arithmetic of elliptic curves) reads this is clear because $M$ has discrete topology. But domain of $f$ is not $M$, so it's not unclear for me. Thank you in advance.
We don't need that $M$ is discrete, it can be any abelian topological group. The map is continuous since it is built from continuous maps. This phrase can be made precise by at least two methods.
Method 1: Elementary reasoning
The action map $w : G \times M \to M$ is continuous by assumption. For every $m \in M$ the map $G \to G \times M$, $g \mapsto (g,m)$ is continuous. It follows that the composition $G \to M$, $g \mapsto gm$ is continuous. The constant map $G \to M$, $g \mapsto -m$ is also continuous. It follows that the sum of these maps $G \to M$, $g \mapsto gm - m$ is also continuous.
Method 2: Category theory
Let us not work with maps of sets, but rather with a category where everything is continuous by default, namely the category of topological spaces and continuous maps. Inside of it, $G$ is just a group object, and $M$ is an abelian group object. Any morphism that we construct in this category is, by definition, a continuous map. So there is no need to verify that something is continuous.
More generally, let $\mathcal{C}$ be any category with finite products. Let $G$ be a monoid object in $\mathcal{C}$ and let $M$ be an abelian group object in $\mathcal{C}$. We define an action of $G$ on $M$ as a morphism $w : G \times M \to M$ satisfying the usual properties. You can express these properties via commutative diagrams, but it is more convenient to apply the Yoneda Lemma, which tells us that they are equivalent to saying that for all test objects $T$ the induced morphism $w(T) : G(T) \times M(T) \to M(T)$ is an action of the monoid $G(T)$ on the abelian group $M(T)$ (in the category of sets). Of course, if $\mathcal{C}$ is concrete with a forgetful functor $\mathcal{C} \to \mathbf{Set}$ preserving finite products, we can alternatively say that $w$ is an action on the underlying sets.
Now for every global element $m : 1 \to M$ we define the morphism $f : G \to M$, using the Yoneda Lemma, by the maps of sets $f(T) : G(T) \to M(T)$ sending $g$ to $ gm - m$. We don't need to verify that $f$ is a morphism - in this setting, we only have morphisms and nothing else.
You can also construct $f$ without the Yoneda Lemma, of course. We define
$a : G \xrightarrow{\cong} G \times 1 \xrightarrow{G \times m} G \times M \xrightarrow{w} M$
$b : G \xrightarrow{\exists !} 1 \xrightarrow{m} M \xrightarrow{-} M$
And then
$f : G \xrightarrow{\Delta} G \times G \xrightarrow{a \times b} M \times M \xrightarrow{+} M$