If $A \le _m B$, then $A$ decidable $\not\Rightarrow B$ decidable
Consider $A = \Sigma^*$ and $B \subset\Gamma^*$ undecidable. If you know any $b \in B$, the constant mapping $\Sigma^* \to \Gamma^*, w \mapsto b$ provides a counterexample.

To be a bit more specific: Let $\Gamma = \{\texttt&lt;, \texttt&gt;, \texttt[, \texttt] \texttt+, \texttt-\}$ be the set of symbols for Brainfuck without I/O. As BF is Turing complete, we know that the subset $B \subset \Gamma^*$ of all terminating BF programs is undecidable (via the Halting problem). It turns out that $\texttt{+} \in B$.

Thus we have $A \leq_m B$ for all languages $A \subseteq \Sigma^*$ via the mapping $f\colon \Sigma^* \to \Gamma^*$ given by $f(w) = \texttt+$.