Is reduction symmetric?
Reduction is not symmetric. Intuitively, if $A$ reduces to $B$, then $B$ is at least as hard as $A$. Just as the usual comparison is not symmetric — for example $3 &gt; 2$ but $2 \not&gt; 3$ — reduction, which is a (partial) order of hardness, isn't symmetric.
For example, SAT reduces to the halting problem, but the halting problem doesn't reduce to SAT.