Let $A$ be a set and $f$ a function with $f:A\to A$

Question

Here are the problems below. My main concern is with part a. I get confused when we at first define $f$ to make to $A\to A$, doesn't this imply that $f$ is onto already? Or is this just a bound given. My other thoughts on part a is below:

a. Suppose $f$ is one to one. Is $f$ then onto?

Im not sure how to prove this, I thought it was true but I was told this was false by using a positive function like $f(x)= e^x$ that maps to $\mathbb R\to \mathbb R$ ($\mathbb R$ is real numbers) and this proves it is not onto since $e^x$ only maps to positive values in $\mathbb R$. But how can we say this is $f:\mathbb R\to \mathbb R$ shouldn't this mean that we need to pick a function that specifically does just that. I mean doesn't $e^x$ map from $\mathbb R\to \mathbb R_+$ (positive $\mathbb R$) not $\mathbb R\to \mathbb R$?

Example that makes sense to me:

b. Suppose $f$ is onto, is $f$ then one to one?

Similar idea with this one, I was told it was false and given an odd function like $F(x)=x^3-x$ with $A: \mathbb R\to \mathbb R$ and told it was onto but not one to one. I understand this reasoning since $f$ maps to all $\mathbb R\to \mathbb R$ but it isn't one to one since $f(-1)=0$ and $f(0)=0$

Answer 1

A function $f: A\to B$ is simply a mapping that, for every element $a\in A$, gives you precisely one element $f(a)\in B$.

There is no requirement that $f$ needs to cover all elements of $B$, in fact, that's the whole point of defining onto functions.

You must realize, however, that a function is not only defined by what it does, but also where it does it.

For example, take these four functions:

• The function $f_1:\mathbb R\to \mathbb R$, defined by $f_1(x) = e^x$
• The function $f_2:\mathbb R \to \mathbb R^+$, defined by $f_2(x)=e^x$.
• The function $g_1:\mathbb R\to \mathbb R$, defined by $g_1(x)=x^2$.
• The function $g_2: \mathbb R \to \mathbb R^+$, defined by $g_2(x)=x^2$.

These functions have different properties:

1. $f_1$ is one-to-one, but not onto.
2. $f_2$ is one-to-one and onto
3. $g_1$ is neither one-to-one nor onto.
4. $g_2$ is not one-to-one, but it is onto.