 Solution

With the convention $\mathbb{N}=\{0,1,2,\dots \}$, the even inputs $2j$ map to $j$ for $j\ge 0$, while the odd inputs $2j+1$ map to $-(j+1)$ for $j\ge 0$. These two images of a function are disjoint and together equal $\mathbb{Z}$. Within either class there are no repeated outputs. More explicitly the inverse function isConsequently $f$ is both injective and surjective, hence a bijection. If instead $\mathbb{N}=\{1,2,\dots \}$, the input zero is absent: $f$ remains injective but its image of a function is $\mathbb{Z}\setminus \{0\}$, so it is not onto. The printed paper does not specify this convention; the zero-based convention gives the apparent intended bijection.