jmc

We can line up the numbers and subtract just as we do in base 10. For example, when we borrow from the $9^1$s place, the digit 1 in the units place becomes $10$, while the digit in the $9^1$s place decreases by 1. Continuing in this way, we find \begin{array}{c@{}c@{\;}c@{}c} & & \cancelto{7}{8} & \cancelto{10}{1}_9 \\ &- & 7 & 2_9 \\ \cline{2-4} & & & 8_9, \end{array}so the difference is $\boxed{8_9}$.