Let $Y = \{n^{2} : n \in N\} \subset N$. Consider $f: N \rightarrow Y$ as $f(n) = n^{2}$. Show that $f$ is invertible. Find the inverse of $f$.

(N/A) To show that $f$ is invertible,we need to find a function $g: Y \rightarrow N$ such that $g \circ f = I_{N}$ and $f \circ g = I_{Y}$.
An arbitrary element $y$ in $Y$ is of the form $n^{2}$ for some $n \in N$.
This implies that $n = \sqrt{y}$.
We define a function $g: Y \rightarrow N$ by $g(y) = \sqrt{y}$.
Now,calculate the compositions:
$(g \circ f)(n) = g(f(n)) = g(n^{2}) = \sqrt{n^{2}} = n = I_{N}(n)$.
$(f \circ g)(y) = f(g(y)) = f(\sqrt{y}) = (\sqrt{y})^{2} = y = I_{Y}(y)$.
Since $g \circ f = I_{N}$ and $f \circ g = I_{Y}$,the function $f$ is invertible.
The inverse function is $f^{-1}(y) = \sqrt{y}$ for all $y \in Y$.