Explanation

(c)
Given, $f(n) = \left\{ \begin{array}{r}
\frac{n - 1}{2},\ when\ n\ is\ odd \\
- \frac{n}{2},\ when\ n\ is\ even
\end{array} \right.\$
And $f:N \rightarrow I$, where $N$ is the set of natural numbers and $I$ is the set of integers.
Let $x,\ y \in N$ and both are even.
Then, $f(x) = f(y)$
$\Rightarrow - \frac{x}{2} = - \frac{y}{2} \Rightarrow x = y$
Again, $x,\ y \in N$ and both are odd.
Then, $f(x) = f(y)$
$\Rightarrow \ \ \ \ \ \frac{x - 1}{2} = \frac{y - 1}{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ x = y$
So, mapping is one-one.
Since, each negative integer is an image of even natural number and positive integer is an image of odd natural number. So, mapping is onto.