Vtyjkyuk

I have Mathematics IV class. I need to prove that inverses of all functions is an one to one function. I cannot do anything. Is there anybody can help me?

$f(x) = y Rightarrow f^-1(x)$ is a one to one function?

Let $f:A to B$ be a function. Suppose that the inverse function $g:B to A$ is noninjective. Sso there exist $b_1 neq b_2 in B$ so that $g(b_1)=g(b_2)$. Then $f$ is not well defined since $f circ g(b_1) neq f circ g(b_2)$, meaning that one element gets mapped to two different places.

This is a failure of the "vertical line test" in elementary calculus if that helps.

Suppose that $f:Xrightarrow Y$ is a bijective function. Then for every $yin Y$ there exists a unique element $x_yin X$ such that $f(x_y)=y$. One can then define a new function $f^-1:Yrightarrow X:ymapsto x_y$. Clearly one has that $fcirc f^-1=Id_Y$ and $f^-1circ f=Id_X$. The goal is to show that $f^-1$ is bijective as well.

Suppose that $f^-1(y)=f^-1(y')$, by applying $f$ to this equalty we get that $y=y'$, hence $f^-1$ is injective. Now choose $xin X$, then $f^-1(f(x))=x$, hence $f^-1$ is surjective.

Let $f : E to F$ be a function. If $f^-1$ exists, then by definition :

$forall x in E, f^-1(f(x)) = x $ and $forall y in F,f(f^-1(y)) = x $

From there you can easily show that both $f$ and $f^-1$ are both "one to one or more" AND "one to one or less" which is "one to one".

Let me give you a little more general statement:

Lemma. Let $fcolon Ato B$ and $gcolon Bto C$ functions. Then

$a)$ if $gcirc f$ is injective, so is $f$,

$b)$ if $gcirc f$ is surjective, so is $g$.

Now, assuming $fcolon Ato B$ is invertible, we have that there exists $gcolon Bto A$ such that $gcirc f = operatornameid_A$, $fcirc g = operatornameid_B$, but then by $a)$ $g$ is injective since $operatornameid_B$ is, which answers your question.

Proof of Lemma. I will prove $a)$ and encourage you to prove $b)$ as an exercise. Let $x,yin A$ such that $f(x) = f(y)$. Then $(gcirc f)(x) = (gcirc f)(y)$ which implies $x = y$ since $gcirc f$ is injective. This proves that $f$ is injective as well.

But, since we are here, let's mention a very useful proposition as well:

Proposition. Function $fcolon Ato B$ is invertible if and only if it is bijective.

Before the proof, let's see how this answers your question. Assume that $fcolon Ato B$ is invertible, i.e. there exists $gcolon Bto A$ such that $gcirc f=operatornameid_A$, $fcirc g = operatornameid_B$. But then $g$ is invertible as well and hence by our proposition bijective.

Proof of the Proposition. Assume that $f$ is bijective. That means that for any $bin B$ there is unique $ain A$ such that $f(a) = b$. Define $gcolon Bto A$ with $g(b) = a$. Then for any $ain A$, $g(f(a)) = a$ by definition of $g$ and we have $f(g(b)) = f(a) = b$, for all $bin B$, i.e. $gcirc f=operatornameid_A$ and $fcirc g = operatornameid_B$. Hence, $f$ is invertible with $g$ its inverse.

For the other direction, use the Lemma.