How to tell if a function is onto or one-to-one I'm practicing what we learned in lecture today and unfortunately I have little to no understanding about the material. I only know the difference of these functions only when a diagram is present (and I can't always have that, so I need to learn how to figure it out without one)
So I've provided an example from my textbook (not assigned work)
Question: Determine whether each of these functions from $\mathbb{Z}$ to $\mathbb{Z}$ is one-to-one
$a$) $f(n)=n-1$ (ANS: onto)
$b$) $f(n)=n^2+1$ (ANS: one-to-one)
I know the answers only since I looked in the back, but have no idea why. Can someone please explain? I will be using the answers as a base to complete the rest of the questions for study. 
 A: One of the answers is wrong. $f(n) = n^2 +1$ is not one-to-one, it is two-to one. (Do you understand what I mean?).  The reason why $f(n) = n-1$ is onto, is because for any integer $m$, the successor integer, $m+1$ corresponds to it. Explicitly, $f(m+1) = (m+1) -1 = m$
A: Let $\mathrm f$:$R\rightarrow S$ be a relation. 
The relation is one-one if every elemt in $R$ is mapped uniquely to every element in S. That means for every x $\in$ R $\exists$ y $\in$S such that
$$f(x)=y$$The relation is then said to be one-one.
Now,the function(a relation which is one one and onto) is said to be onto if every element of $R$ is mapped to every element in $S$. To determine whether the function is one-one and onto the domain and range of the function has to be known
As an example lets take the function
$$y=x^2 \text{, x $\in R$}$$
The relation here is not one-one as you have 2 different values of x which gives the same value of y.$$f(-2)=f(2)=4$$
However if we restrict the domain of the function to the set of positive natural numbers(or negative) the function becomes one-one.
A: A function $f:A\rightarrow B$ is one-to-one if whenever $f(x)=f(y)$, where $x,y \in A$, then $x=y$. So, assume that $f(x)=f(y)$ where $x,y \in A$, and from this assumption deduce that $x=y$. 
If we assume that $f(n_1)=f(n_2)$ where $n_1,n_2\in \mathbb{Z}$, that is we assume that $(n_1)^2+1=(n_2)^2+1$, then $(n_1)^2=(n_2)^2$. This does not imply that $n_1=n_2$, consider $n_1=1$ and $n_2=-1$. Thus $f(n)=n^2+1$ is not a one-to-one function.
