# Inverse function on given sets

My question is: Given sets $A$ = {$a_1, a_2$} and B = {$b_1$}, let the function $f$ from $A$ to $B$ be given by the following set of ordered pairs, $f$ = { ($a_1, b_1$), ($a_2, b_1$) }. If $f$ has an inverse function, call it $g$, and write $g$ as a set of ordered pairs. If $f$ does not have an inverse function, explain why it doesn’t.

I was thinking that there cannot be an inverse function because the domain of the inverse function, $g$ = {($b_1, a_1$), ($b_1, a_2$)}, is $b_1$, which is the same and is pointing to different ranges. For a function to exist the domain has to be unique correct?

• Your reasoning is correct but your wording is strained. "The domain of the inverse function g"=> g is not the inverse function and g is not a function at all.. The domain of the relation g is b1. "which is the same" The same as what? "Pointing to different ranges". It's a single value pointing to two values so so the range is {a1,a2} is a single range. Not two different ranges. "the domain has to be unique". {b1} is unique. Dec 11 '15 at 17:43
• Better wording might be. a1 and a2 are different values in the domain of f yet f(a1) and f(a2) are the same value b1. (Such a function by definition is not injective.) Such a function can not have an inverse, g, as g(b1) would be defined to be the value x such that f(x) = b1. As there are two such values, a1 and a2 the definition is ambiguous. (And if we specified one over the other, say g(b1) = a1. Then g(f(a2)) = a1 and not a2 failing the condition that an inverse does invert all values.) Dec 11 '15 at 17:53

Your reasoning is basically correct (you need to assume $a_1\neq a_2$ though!)
Your wording is a little bit off though, I believe. The range is the set of all possible values a function has. A single value (like $a_1$ or $a_2$) is just called, well, a value. For example, the range of $f$ is $B$. The domain of a function is again a set: the set of all possible "inputs". In your case, the domain of $f$ is $A$. Calling $g$ an inverse function is a bit troublesome, since it is of course not a function. You can call $g$ the inverse relation of $f$. The convention "Function = Graph" makes every function surjective (onto) by default. So eventually, you want to define functions in the following way: $$f = (A, \{ (a_1,b_1),(a_2,b_1) \}, B)$$ or similarly. Here $f$ is surjective so it does not actually make a difference, but if you had $B = \{b_1,b_2\}$ with $b_1\neq b_2$ it would matter.
If $f$ has an inverse, then $f$ is bijective, hence injective (one-to-one), but $f(a_1)=f(a_2)$ even though $a_1\neq a_2$ (Contradiction).