If A is mapping reducible to B then the complement of A is mapping reducible to the complement of B
As Dave said, it follows from a simple logical equivalence: $p \leftrightarrow q$ is the same as $\lnot p \leftrightarrow \lnot q$. Now put $p= w\in A$ and $q = f(w) \in B$.

$A \leq_m B$ means there is a total computable $f$ s.t. for all $w$,


  $w \in A \leftrightarrow f(w) \in B$. 


By the argument above, this is the same as


  $w \notin A \leftrightarrow f(w) \notin B$. 


Or equivalently


  $w \in \bar{A} \leftrightarrow f(w) \in \bar{B}$. 


And therefore, the same $f$ shows that $\bar{A} \leq_m \bar{B}$.