I'm reading up on how to prove if a function (represented by a formula) is one-to-one or onto, and I'm having some trouble understanding.
To prove if a function is one-to-one, it says that I have to show that for elements $a$ and $b$ in set $A$, if $f(a) = f(b)$, then $a = b$. I understand this to mean that if two elements in a domain map to the the same element in a codomain, then for the function to be one-to-one, they must be the same element because by definition, a one-to-one function has at most one element in the domain mapped to a particular element in the co-domain. Did I understand this correctly?
Then to prove that the function is onto, I'm reading an example that says "let's prove that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 5x+2$ is onto, where $\mathbb{R}$ denotes the real numbers. We let $y$ be a typical element of the codomain and set up the equation $y =f(x)$. then, $y = 5x+2$ and solving for $x$ we get $x ={y-2\over 5}$. Since $y$ is a real number, then ${y-2\over 5}$ is a real number and $f({y-2\over 5})=5({y-2\over 5})+2=y.$
I'm not really seeing how that proves anything, so can anybody explain this to me?
Best Answer
Yes, your understanding of a one-to-one function is correct.
A function is onto if and only if for every $y$ in the codomain, there is an $x$ in the domain such that $f(x) = y$.
So in the example you give, $f:\mathbb R \to \mathbb R,\quad f(x) = 5x+2$, the domain and codomain are the same set: $\mathbb R.\;$ Since, for every real number $y\in \mathbb R,\,$ there is an $\,x\in \mathbb R\,$ such that $f(x) = y$, the function is onto. The example you include shows an explicit way to determine which $x$ maps to a particular $y$, by solving for $x$ in terms of $y.$ That way, we can pick any $y$, solve for $f'(y) = x$, and know the value of $x$ which the original function maps to that $y$.
Side note:
Note that $f'(y) = f^{-1}(x)$ when we swap variables. We are guaranteed that every function $f$ that is onto and one-to-one has an inverse $f^{-1}$, a function such that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.