How to prove $C^0(G,A)$ isomorphic to $A$

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Let $G$ be a profinite group and $A$ an abstract $G$-module. Let $C^0(G,A)$ be the abelian group of all continuous maps from $G$ to $A$ fixed by the action of $G$. The map which sends $x \in C^0(G,A)$ to $x(1)$ gives a natural isomorphism of the two groups apparently... I can prove that the map is injective but I am having trouble proving it is surjective. Could someone please explain me how this works? Thank you very much.

PS Given $\sigma \in G$ it acts of the map $x: G \rightarrow A$ by $(\sigma x) (\cdot)= \sigma x (\sigma^{-1} \cdot)$.

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For surjectivity, let $a \in A$, then we can define $f_a: G \to A$ by $f_a(g)=ga$.

This is continuous and $G$-linear as $A$ is a topological $G$ module and we have $f_a(1)=a$. The fact that it is fixed by the action of $G$ on $C^0(G,A)$ follows from the $G$-linearity.