Functions images and inverse images

Question

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?

Answer 1

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?

Answer 2

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. $$k = -1$$ $${ {1} \times {-1} }\cup{ {-1} \times {-1} }$$

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

Answer 3

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\}$

$= \{n^2|n\in \mathbb Z\}$

$= \{0,1,4,9,.....\}$