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?
Best Answer
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?