By a Borel measurable mapping $f$ you must mean that $f$ is a Borel map. (That is, I guess you're not assuming that there are any measures floating around.) In that case, the answer is no:
Let $f$ be the identity function $\mathbb{N}_d \to \mathbb{N}_i$, where $\mathbb{N}_d$ has the discrete topology and $\mathbb{N}_i$ has the indiscrete (or trivial) topology. Then $f$ is surjective, but its inverse is not Borel. For instance, $f^{-1}\{3\} = \{3\}$ is not a Borel set in $\mathbb{N}_i$, though $\{3\}$ is Borel in $\mathbb{N}_d$. (Of course, there's nothing special about $\mathbb{N}$ here: pick your favorite set with at least two elements.)