\(f\) is not injective. To prove this, we must simply find two different elements of the domain which map to the same element of the codomain. Since \(f(\{1\}) = 1\) and \(f(\{2\}) = 1\text{,}\) we see that \(f\) is not injective.
\(f\) is not surjective. The largest subset of \(A\) is \(A\) itself, and \(|A| = 10\text{.}\) So no natural number greater than 10 will ever be an output.
\(f\inv(1) = \{\{1\}, \{2\}, \{3\}, \ldots \{10\}\}\) (the set of all the singleton subsets of \(A\)).
\(f\inv(0) = \{\emptyset\}\text{.}\) Note, it would be wrong to write \(f\inv(0) = \emptyset\) - that would claim that there is no input which has 0 as an output.
\(f\inv(12) = \emptyset\text{,}\) since there are no subsets of \(A\) with cardinality 12.