If A is polynomial time reducible to B does that imply not(A) is polynomial reducible to not(B)
In case you were wondering, the answer is yes. 

Here is the reasoning.

Lemma: Let $f$ be a reduction from $A$ to $B$. Then $f$ is also a reduction from $\overline A$ to $\overline B$.
Proof: By the definition of a reduction, $f$ maps $A$ to $B$ and $f$ maps the complement of $A$ to the complement of $B$. So $f$ maps the complement of the complement of $A$, which is $A$ itself, to $B$, which is the complement of the complement of $B$. By the definition of a reduction again, $f$ is a reduction from the complement of $A$ to the complement of $B$. QED.







Suppose $A$ is polynomial-time reducible to $B$. Let $f$ be a polynomial reduction from $A$ to $B$. The lemmas shows that $f$ is also a reduction from the complement of $A$ to the complement of $B$. Since $f$ is polynomial-time, $\overline A$ is polynomial-time reduced to $\overline B$ by $f$.

With $A$ replaced by $\overline A$ and $B$ by $\overline B$, the above paragraph also shows $\overline A$ is polynomial-time reducible to $\overline B$ implies $A$ is polynomial-time reduced to $B$.

The above two paragraphs shows the equivalence of $A\leq_p B$ and $\overline{A}\leq_p \overline{B}$.