The problem \textsc{Token Jumping} asks whether, given a graph $G$ and two independent sets of \emph{tokens} $I$ and $J$ of $G$, we can transform $I$ into $J$ by changing the position of a single token in each step and having an independent set of tokens throughout. We show that there is a polynomial-time algorithm that, given an instance of \textsc{Token Jumping}, computes an equivalent instance of size $O(g^2 + gk + k^2)$, where $g$ is the genus of the input graph and $k$ is the size of the independent sets.