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I am aware there are great answers in

Inverse of a Function exists iff Function is bijective

I was wondering if I could do this proof directly.

My plan is to prove the given statement for all relations (because to me arbitrary relations seem nicer than actual functions), so that it holds for functions as well. Like so,

Let $R \subseteq A\times B$ be a relation. Assume its inverse relation $R^{-1}$ is a function. Suppose $a_1 R b$ and $a_2Rb$. Then by definition of inverse relations, $bR^{-1}a_1$ and $bR^{-1}a_2$. Since $R^{-1}$ is a function, $b$ must have a unique image under $R^{-1}$ meaning $a_1 = a_2$. This proof would've worked if $R$ was a function.

Since the plan above failed(?), I will try to adopt the general idea to a proof where $R$ is a function. So,

Consider the function $f: A \to B$. Let $f^{-1}$ be a function. Suppose $f(a_1) = f(a_2) = y \in B$. Since $f$ is a relation, I can use the notation for relations and so $a_1fy$ and $a_2fy.$ By definition of inverse relations, $yf^{-1}a_1$ and $yf^{-1}a_2$. Since $f^{-1}$ is a function, $y$ must have a unique image under $f^{-1}$ meaning $a_1 = a_2$. This proof looks like it works, but I have never seen anyone argue like this or use this notation.

I'd be happy if the second proof works despite the fact it's ugly because it would mean everything I learned so far I understood correctly and I am not going to show this to anyone (just for personal use). So, is the second proof in/correct? If incorrect, how can I fix it while keeping the main idea in place? Thanks.

edit: I want to prove "If $f : A \to B, g : B \to A$ are functions with $g \circ f = i_A$ and $f \circ g = i_B$, then $f, g$ are bijective and $g = f^{−1}$" from the original statement. I cannot assume this theorem.

fgh
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  • And... how do you define the inverse of a function? – azif00 Oct 15 '19 at 01:12
  • The relation $f^{-1}$ is an inverse function if the function $f$ is bijective. – fgh Oct 15 '19 at 01:15
  • So, you can only follow the implications: $$f^{-1}\textrm{ is a function} \ \Leftrightarrow \ f\textrm{ is bijective} \ \Rightarrow \ f\textrm{ is injective}$$ – azif00 Oct 15 '19 at 01:28
  • What is the benefit of using the relation notation? $f$ is a function hence it is a relation . As you have pointed out we can have $f^{-1}$ as a function where $f$ is a relation. As $f^{-1}=x^2$ .. so why you want to use the relation notation where this fact you want to prove is correct in case of functions not general relation – IrbidMath Oct 15 '19 at 01:41

1 Answers1

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In order to prove that $f$ is injective you show that $$f(x)=f(y)\implies x=y$$.

Let $f(x)=f(y)$, since $f^{-1}$ is a function you have $$f^{-1} ( f(x))= f^{-1} (f(y))$$ That is $x=y$ so $f$ is injective.

Mohammad Riazi-Kermani
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  • I am sorry, I forgot to say I am not assuming composition of inverse functions that give identity because I was trying to prove a related statement from the one given in the OP. – fgh Oct 15 '19 at 00:56