First some definitions:
Let $G$ be a group. Let $F$ be the family of all subgroups of finite index. Let $\{x_n\}$ be a sequence in $G$. We define this sequence to be Cauchy if given $H\in F$ there exists $n_0$ such that for $m,n\geq n_0$ we have $x_nx_m^{-1}\in H$. We define multiplication of two sequences component-wise. I have already proven that the set $C$ of Cauchy sequences forms a group.
Next, define $\{x_n\}$ to be a null sequence if given $H\in F$ there exists $n_0$ such that for $n\geq n_0$ we have $x_n\in H$. I have already proven also that the null sequences $N$ form a normal subgroup of the group of Cauchy sequences. Finally, the factor group of all Cauchy sequences modulo the null sequences is called the completion of $G$.
Now the question: Prove that the map which sends an element $x\in G$ on the class of the sequence $(x,x,\ldots ,x)$ modulo null sequences is a homomorphism of $G$ into the completion, whose kernel is the intersection $\bigcap_{H\in F}H$.
EDIT: I can already prove myself that it is a homomorphism but I'm struggling with the the kernel.
Well, literally, this map $x\mapsto (x,x,x,\dots)$ is rather a map $G\to C$, but we can compose it with the canonical $C\to C/N$, and this is what is meant here.
The [coset of] $\ (x,x,x,\dots)$ becomes the identity in $C/N$ iff $(x,x,x,\dots)\in N$ which now means $x\in H$ for all $H\in F$.