In the definition of contraction mapping, it requires:
$(1)$ the function $f$ must map the domain $A$ to $A$, where $A\subseteq\mathbb{R}$.
$(2)$ There exists a constant $0<c<1$, such that $\forall x,y\in A, |f(x)-f(y)|\le c|x-y|$
My question is, why it must require $(1)$, and is there a counter-example such that $f: A\to B$, where $A\neq B$, but it is not a contraction mapping? To make this counter-example more strong, assume both $A$ and $B$ are compact sets.
In this post, it shows $\cos(x)$ is a contraction mapping on $[0, \pi]$, but clearly it is not a $A\to A$ mapping.
The general idea of a contraction mapping is that it shrinks the distance between any two points. It is easy to generalize the idea, let $f: (A,d_a) \rightarrow (B,d_b)$, f is a contraction if $d_b(fx, fy) \leq \alpha d_a(x,y)$ for some $\alpha < 1$.
Since you mention compactness, it is possible that you are discussing this in the context of fixed point theorem. But if $(A,d_a), (B,d_b)$ share no points, then it is impossible that there will be any $x$ such that $f(x) = x$.