Note that \(g(1) \ne g(\{1\})\text{.}\) The first is an element: \(g(1) = 2\text{.}\) The second is a set: \(g(\{1\}) = \{2\}\text{.}\)
To find \(g\inv(1)\text{,}\) we need to find all integers \(n\) such that \(n^2 + 1 = 1\text{.}\) Clearly only 0 works, so \(g\inv(1) = \{0\}\) (note that even though there is only one element, we still write it as a set with one element in it).
To find \(g\inv(2)\text{,}\) we need to find all \(n\) such that \(n^2 + 1 = 2\text{.}\) We see \(g\inv(2) = \{-1,1\}\text{.}\)
Finally, if \(n^2 + 1 = 3\text{,}\) then we are looking for an \(n\) such that \(n^2 = 2\text{.}\) There are no such integers so \(g\inv(3) = \emptyset\text{.}\)