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?

  • $\begingroup$ Welcome to MSE! I edited your question, but I don't know what you mean with $f^{-1}(z*(1))$. Can you elaborate on that? $\endgroup$ – user370967 Jun 20 '17 at 16:40
  • $\begingroup$ I meant f^-1 (z x {1}) I believe its a cross product of integer and the element 1, I wasn't able to add the curly braces as mutex was considering it as small braces. $\endgroup$ – Prathik Gurudatt Jun 20 '17 at 16:49
  • $\begingroup$ Ah good. Look at the code of my edit to see how to do that :) $\endgroup$ – user370967 Jun 20 '17 at 16:50
  • $\begingroup$ $f^{-1}(\Bbb Z\times \{1\})$ makes no sense. $\Bbb Z\times \{1\}\subseteq \Bbb Z^2$, hence a subset of the domain and not the codomain. But $f^{-1}$ takes elements of $\Bbb Z$, the codomain of $f$. $\endgroup$ – M. Winter Jun 20 '17 at 17:27
  • $\begingroup$ I believe last point should be $f(\Bbb Z\times \{1\})$ ? $\endgroup$ – Archetype2142 Jun 20 '17 at 21:53

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


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

Can you proceed now?

  • $\begingroup$ So are u saying that, if the f(n,k) = 0? we must find the values of n and k that converge it to 0? There could be multiple cases or multiple answers? does it require permutations? $\endgroup$ – Prathik Gurudatt Jun 20 '17 at 16:52
  • $\begingroup$ Wow you seem to be very confused. What is the definition of $f^{-1}(B)$ where $B$ is a subset of the codomain? They ask you to find this set for $B = \{0\} \subset \mathbb{Z}$ $\endgroup$ – user370967 Jun 20 '17 at 16:58
  • $\begingroup$ You have written that "find the pairs (n,k) such that f(n,k) = 0" so are you saying that we have to find the values of n and k that leads to 0? I'm sorry I really am confused, please bear with me, that is what brought me here. But i do have some partial clarity where your heading to. $\endgroup$ – Prathik Gurudatt Jun 20 '17 at 17:03
  • $\begingroup$ for a function $f: a \to b, f^{-1}(B) := \{a \in A| f(a) \in B\}$ so for $B = \{0\}$, we have $f^{-1}(\{0\}) := \{a \in A| f(a) \in \{0\}\} = \{a \in A| f(a) =0\}$, here $A = \mathbb{Z}^2$ $\endgroup$ – user370967 Jun 20 '17 at 17:08
  • $\begingroup$ @PrathikGurudatt This task has nothing to do with convergence. You should look up how the preimage $f^{-1}$ is defined. $\endgroup$ – M. Winter Jun 20 '17 at 17:32

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\}$

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

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

  • $\begingroup$ Good detailed answer. +1, but you shouldn't give a full answer $\endgroup$ – user370967 Jun 21 '17 at 7:02

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

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

  • $\begingroup$ 0.1 is in Z x N but f(0,1) = 0 not in N so f^{-1}(N) != Z x N. $\endgroup$ – fleablood Jun 20 '17 at 23:01
  • $\begingroup$ What does "${0}\times\mathbb{Z} \lor \mathbb{Z} \times {0}"$ mean? What does $\lor$ mean in set notation. I thought $\lor$ was a logical operator for "OR". ${0}\times\mathbb{Z} $ OR $ \mathbb{Z} \times {0}$ is not a set, is it? $\endgroup$ – fleablood Jun 20 '17 at 23:03
  • $\begingroup$ If $n$ is not a perfect square then $n \not \in f(\mathbb Z \times 1)$. $\endgroup$ – fleablood Jun 20 '17 at 23:08
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    $\begingroup$ Waw this illustrates perfectly how this site malfunctions. The OP clearly not understands a concept. He gets spoonfed the answer and accepts it because someone else made his homework. Hooray! $\endgroup$ – user370967 Jun 21 '17 at 7:01
  • $\begingroup$ @Ritwick You should really stick to mathematical standard notation, i.e. $1.2\to1\cdot 2$, $\vee\to\cup$ for sets and $1\times-1\vee-1\times 1$ does not make any sense but should certainly mean $\{(1,-1),(-1,1)\}$. Especially for an OP as fleablood who seems to have no clue here and should really not copy paste this bad/wrong notation. $\endgroup$ – M. Winter Jun 21 '17 at 10:17

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