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   $ f^{-1}(\{0\}) $,  $ f^{-1}(\mathbb{N}) $  and  $  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?
 A: 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?
A: For any $g: A \rightarrow B$ and $C \subset B$, then by definition $g^{-1}(C) := \{x \in A| f(x) \in C\}$.  Conceptually $g^{-1}(C)$ is simply the set of all points originally in the domain that map into $C$.
1) So $f^{-1}(0)$ is simply all the pairs that map to $0$ or $\{(n,k)|f(n,k) = n^2 k = 0;n,k \in \mathbb Z\}$. 
So if $n^2 k = 0$ either $k = 0$ or $n = 0$ so $f^{-1}(0) = \{(n,k)| n = 0 \text{ or } k = 0\} = \{(n,0)|n \in \mathbb Z\} \cup \{(0,n)| n \in \mathbb Z\} = [\mathbb Z \times \{0\}] \cup [\{0\}\times \mathbb Z]$
2) $f^{-1}(\mathbb N)$ are all the paris that map to a natural number (positive integer) or $\{(n,k)|f(n,k) = n^2k > 0; n,k \in \mathbb Z\}$.
So if $n^2k > 0$ then either $n^2 > 0$ and $k> 0$ or $n^2 < 0$ and $k < 0$.  $n^2 < 0$ is impossible so $n^2 > 0$ and $k > 0$ and therefore $n \ne 0$.
So  $f^{-1}(\mathbb N) = \{(n,k)| n \ne  0,k > 0; n,k \in \mathbb Z\} = \{(n,k)|n \in \mathbb Z \setminus\{0\}, k \in \mathbb N\} = [\mathbb Z\setminus \{0\}]\times \mathbb N$.
3)The last one is not the pre-image but the (mapped to) image.
$g(C) = \{g(c)|c \in C\}$
So $f(\mathbb Z \times 1\}= \{ f(n,k)|(n,k) \in \mathbb Z \times \{1\}\}$
$ = \{n^2k|n\in \mathbb Z; k = 1\}$
$= \{n^2|n\in \mathbb Z\}$
$= \{0,1,4,9,.....\}$
A: 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.
$$k = -1$$
$${ {1} \times {-1} }\cup{ {-1} \times {-1} }$$

*$f(\mathbb{Z}\times{1})$
$$n^2k = integer^2 \times 1 = natural$$
