I remenber here the concept of a contraction mapping.
Definition: Let (X, d) be a metric space. Then a map T : X → X is called a contraction mapping on X if there exists q ∈ [0, 1) such that
$$ d(T(x),T(y))\le qd(x,y) d(T(x),T(y))\le q d(x,y) \ \ x, y \in X.$$
Assume that, $f :B \longrightarrow B$ is a contraction mapping and $B$ is a closed ball. I'd like to know if always $$ f(\partial B)=\partial (f(B)) ? $$
The answer is no. As a counterexample consider $f(x_1, \dots, x_n) = (|x_1|, x_2, \dots, x_n)$ on the unit ball. Then $f(\partial B) = \{x \in S^{n-1} : x_1 \ge 0\}$ and $\partial f(B) = f(\partial B) \cup \{x \in B: x_1 = 0\}$.