= Solution
With the convention $\mathbb N=\{0,1,2,\ldots\}$, the even inputs $2j$ map to $j$ for $j\geq0$, while the odd inputs $2j+1$ map to $-(j+1)$ for $j\geq0$. 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> is
$$
\boxed{f^{-1}(z)=\begin{cases}2z,&z\geq0,\\-2z-1,&z<0.\end{cases}}
$$
Consequently $f$ is both <injective> and <surjective>, hence a <bijection>. If instead $\mathbb N=\{1,2,\ldots\}$, 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>.
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