Let f : Z→Z be given by f(x)

Injectivity:
Let x and y be two elements in the domain (Z), such that

$f(x)=f(y)$

Case-1: Let both $x$ and $y$ be even.

Then,

$f(x)=f(y)$

$\Rightarrow \frac{x}{2}=\frac{y}{2}$

$\Rightarrow x=y$

Case-2: Let both $x$ and $y$ be odd.

Then,

$f(x)=f(y)$

$\Rightarrow 0=0$

Here, we cannot determine whether $x=y$.

So, $f$ is not one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that

Co-domain of $f=Z=\{0, \pm 1, \pm 2, \pm 3, \pm 4, \ldots\}$

Range of $f=\left\{0,0, \frac{\pm 2}{2}, 0, \frac{\pm 4}{2}, \ldots\right\}=\{0, \pm 1, \pm 2, \ldots\}$

$\Rightarrow$ Co-domain of $f=$ Range of $f$

$\Rightarrow f$ is onto.

So, the answer is (a).