We are given the relation $R$ between $\mathbb{N}$ and $\mathbb{N}$: $$ R\colon \mathbb{N}\rightarrow \mathbb{N}\colon n \mapsto n^3- 3n^2 - n $$ and we are asked: is $R$ a map, bijection, injection or surjection?
First of all, it is clear that not every natural number $n$ has a corresponding mapping, e.g. 1 would map to -3, which is $\notin \mathbb{N}$, similarly for 2 (which would map to -6) and 3 (would map to -2). In conclusion: not every element in the source set $\mathbb{N}$ has an associated target element, and also not every element in the target set $\mathbb{N}$ has an associated source element.
This rules out bijection and surjection. It should be an injection, since every target element has a unique source element. My colleague thinks that $R$ is both a map and an injection, since "all injections are maps". However, I believe that $R$ is not a map, since not every element in the source set has a target element. Various sources on the internet (and books) seem to define maps and injections as being restricted to the domain of $R$, in which case my colleague is correct that injections are maps.
We are stuck. Which one of us is right?
EDIT We are following the (non-standard?) definitions from a course on Discrete Mathematics, which are as follows:
- A relation between two sets is called a mapping if from every element there departs exactly one arrow.
- A relation is called an injection if from every element at most one arrow departs and in every element at most one arrow arrives.
A (binary) relation between sets $X$ and $Y$ is nothing else than a subset of the Cartesian product $X × Y$. In your question we have $R \subset \mathbb N \times \mathbb N$. In my opinion it is misleading to write it the form $n \mapsto n^3- 3n^2 - n $, I would prefer to write $$R = \{ (n,m) \in \mathbb N \times \mathbb N \mid m = n^3- 3n^2 - n \} .$$
I recommend to have a look at https://en.wikipedia.org/wiki/Binary_relation.
$R$ is not a function $\mathbb N \to \mathbb N$. As you have shown, for $n = 1, 2,3$ we do not have $m \in \mathbb N$ such that $(n,m) \in R$. However, we may regard it is as a partial function which means that for each $n \in \mathbb N$ we have at most one $m \in \mathbb N$ such that $(n,m) \in R$. The restriction $\mathbb N \setminus \{1, 2, 3\} \to \mathbb N$ is a function.
$R$ is an injective relation. This means that if $(n, m) \in R$ and $(n', m) \in R$, then $n = n'$. In fact, consider the function $\phi : \mathbb R \to \mathbb R,\phi(x ) = x^3 -3x^2-x$. Its derivative $\phi'(x) = 3x^2 - 6x -1$ is positive for $x > \xi = 1 + \sqrt{4/3}$, thus $\phi$ is strictly increasing on $(\xi,\infty)$. Since $ 4 > \xi$, we get injectivity.
$R$ is not a surjective relation which means that there exists $m \in \mathbb N$ such for all $n \in \mathbb N$ we have $(n,m) \notin R$. In fact, we have $\phi(4)= 12, \phi(5) = 45$. Since $\phi$ is strictly increasing on $(\xi,\infty)$, we may take $m = 13$.