Functions images and inverse images The objective of this question is to find if the function is a bijective function or not and if it is a bijective find its images and inverse images. 

$$ f:\mathbb{Z^2} \to \mathbb{Z}$$ 
$$ f(n,k) = n^2k $$

We have to find inverses of &nbsp; $ f^{-1}(\{0\}) $,&nbsp;&nbsp;$ f^{-1}(\mathbb{N}) $&nbsp; and &nbsp;$  f(\mathbb{Z} \times \{1\}) $    

But I fail to understand the approach to this problem, I do understand that they need to have unique mappings and co-domains must be matched, but could anyone help me make it analogous to this situation?

questions such $$y = x^2 $$ is not bijective since they have multiple images and are not bijective. Their inverse will be a sqaure root with + and - and hence its an invalid case. Could someone please  correct my approach?
We have

$$ f:\mathbb{Z^2} \to \mathbb{Z}$$ 
$$ f(n,k) = n^2k $$

This function is obviously not bijective. Many elements get mapped to zero. For example, $f(1,0) = f(-1,0) = 0$ (not injective). The function is surjective though.

So this means that $f$ does not have a well defined inverse function, as bijectivity is required for that. 

However, we can consider the inverse image, even when the inverse function does not exist. You are asked to find $$f^{-1}(\{0\})$$

or in other words: 

Find the pairs $(n,k)$ such that $f(n,k) = 0$. 

So $f(n,k) = 0 \iff n^2k = 0 \iff n = 0 \lor k = 0$

Hence:

$$f^{-1}(\{0\}) = \{(n,k)|n = 0 \quad \mathrm{or} \quad k = 0\} = \{(n,0)|n \in \mathbb{Z}\} \cup \{(0,k)|k \in \mathbb{Z}\}$$

Can you proceed now?
We have 
$$f:\mathbb{Z^2} \to \mathbb{Z}$$
$$f(n,k) = n^2k$$


For Surjection, 
$$n^2k = n.n.k$$
example
$$2 . 2. 2 = 16$$ $$and$$ $$1.1.16 = 16$$
both map to the same number in the codomain, 
Since surjective function is a function whose image is equal to it's codomain it is a surjective function.
For injection,  let's assume
$$n^2k = p^2q$$
example $$2.2.9 \neq 3.3.4$$
Hence, the function is not injective 
$f^{-1}(\mathbb{N})$
$$n^2k \in \mathbb{N}$$
$$= \mathbb{Z} \times \mathbb{N}$$
$f^{-1}(0)$
$$n^2k = 0$$
$${n = 0} \cup {k = 0}$$
$${ {0}\times\mathbb{Z} }\cup{ \mathbb{Z} \times {0} }$$
$f^{-1}({-1})$
$$n^2k = -1$$
$${n = 1} \cup{n = -1}$$ 
Since n is squared, it can either be positive or negative, will always give +1.
$$k = -1$$
$${ {1} \times {-1} }\cup{ {-1} \times {-1} }$$
$f(\mathbb{Z}\times{1})$
$$n^2k = integer^2 \times 1 = natural$$