I've been going back and forth with this for a dew days, can't quite be sure. When I look at the pumping lemma method, I think a context-free language could possibly be reduced to a regular language. 
If a language is m-reducible to a regular language, does it mean that this language is also necessarily a regular language?
complexity theory;formal languages;reductions
The answer to your question is no. There's a general result you can use in this case:Theorem. If $A$ and $B$ are any decidable languages and $B\ne\varnothing$ and $B\ne\Sigma^*$, then $A\le_\text{M} B$.This says that any nontrivial decidable languages are mapping reducible to each other. In other words, mapping reducibility tells you nothing about decidable languages.Here's an example. Let $L=\{0^n1^n\mid n\ge 0\}$. This is a well-known non-regular language. Now let $R$ be, say the regular language $\{0\}$. Then $L\le_\text{M} R$ by the mapping $f:\{0, 1\}^*\rightarrow\{0, 1\}^*$ defined by$$f(s) = \begin{cases}0\quad\text{if $s=0^n1^n$ for some $n\ge 0$}\\1\quad\text{otherwise}\end{cases}$$It's obvious that $f$ is Turing-computable and that $s\in \{0^n1^n\} \Longleftrightarrow f(s)\in R$, which is exactly what you need to show that this non-regular language is mapping reducible to this regular language. 