First, \(g(n) = n - (-1)^n = \begin{cases} n-1 & \text{if } n \text{ is odd} \\ n+1 & \text{if } n \text{ is even} \end{cases}\). For odd n, \(g(n) = n-1\) is even, and for even n, \(g(n) = n+1\) is odd. Now \((f \circ g)(n) = f(g(n))\). If n is odd: \(g(n) = n-1\) (even), so \(f(g(n)) = \frac{n-1}{2}\). If n is even: \(g(n) = n+1\) (odd), so \(f(g(n)) = \frac{(n+1)+1}{2} = \frac{n+2}{2}\). For injectivity: if \((f \circ g)(n_1) = (f \circ g)(n_2)\), we need to show \(n_1 = n_2\). If both are odd: \(\frac{n_1-1}{2} = \frac{n_2-1}{2}\), so \(n_1 = n_2\). If both even: \(\frac{n_1+2}{2} = \frac{n_2+2}{2}\), so \(n_1 = n_2\). If one odd and one even: \(\frac{n_1-1}{2} = \frac{n_2+2}{2}\), giving \(n_1 = n_2+3\), which contradicts parity. So fog is injective. For surjectivity: any \(m \in \mathbb{N}\) can be written as \(m = \frac{2m-1}{2}\) (from odd \(n=2m-1\)) or \(m = \frac{2m+2}{2}\) (from even \(n=2m-2\)). So fog is surjective. Thus fog is bijective.