I'm currently reading up a bit on Turing decidability reductions as I'll take a more in-depth course about the topic next semester. While reading up on some older course material I asked myself the following question which leaves me a bit puzzled:Let $A, B$ be languages, and $\le_m$ be the mapping reduction for languages.For mapping reductions a central statement is If $A \le _m B$, then $B$ decidable $\implies A$ decidableI'm now wondering about is whether the implication also works the other way around:If $A \le _m B$, then $A$ decidable $\implies B$ decidableI strongly suspect that this is not the case as $A \le _m B$ does not even imply the existence of a mapping function from $B$ to $A$, however I'm struggling to find a good counterexample for this. 
If $A \le _m B$, then $A$ decidable $\not\Rightarrow B$ decidable
turing machines;reductions;undecidability
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 $\Gamma = \{\texttt<, \texttt>, \texttt[, \texttt] \texttt+, \texttt-\}$ 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 $f(w) = \texttt+$.