Let $T:\mathbb R^n \to \mathbb R^n $ be an isometry and $T(0)=0$ , then $T$ is linear and $T(B[0,1])\subseteq B[0,1]$ so $T:B[0,1]\to B[0,1]$ is an isometry and since $B[0,1]$ is compact so $T|_{B[0,1]}$ is surjective ( It is well known that if $K$ is a compact metric space and $f:K \to K$ is isometry then $f$ is surjective ) i.e. $T(B[0,1])=B[0,1]$ , so then $T$ is surjective . My question is , suppose we drop the assumption $T(0)=0$ , Is $T$ still surjective ?
Please help . Thanks in advance
You already know that $T$ is linear if $T(0) = 0$. However, consider the maps $$ S(x) = T(x) - T(0)\\ R(x) = x + T(0) $$ Then clearly both $R$ and $S$ are isometries with $T = R \circ S$. However, since $S(0) = 0$, $S$ is linear, and by the rank-nullity theorem any injective linear map is surjective. Moreover, the translation $R$ is clearly surjective. Thus, the composition $R \circ S = T$ must also be surjective.
So, $T$ is indeed necessarily surjective.