Show that the inverse of function $f$ is one to one

Question

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?

Answer 1

Suppose that $f:X\rightarrow Y$ is a bijective function. Then for every $y\in Y$ there exists a unique element $x_y\in X$ such that $f(x_y)=y$. One can then define a new function $f^{-1}:Y\rightarrow X:y\mapsto x_y$. Clearly one has that $f\circ f^{-1}=Id_Y$ and $f^{-1}\circ f=Id_X$. 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 $x\in X$, then $f^{-1}(f(x))=x$, hence $f^{-1}$ is surjective.

Answer 2

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".

Answer 3

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.

Answer 4

Let me give you a little more general statement:

Lemma. Let $f\colon A\to B$ and $g\colon B\to C$ functions. Then

$a)$ if $g\circ f$ is injective, so is $f$,

$b)$ if $g\circ f$ is surjective, so is $g$.

Now, assuming $f\colon A\to B$ is invertible, we have that there exists $g\colon B\to A$ such that $g\circ f = \operatorname{id}_A$, $f\circ g = \operatorname{id}_B$, but then by $a)$ $g$ is injective since $\operatorname{id}_B$ is, which answers your question.

Proof of Lemma. I will prove $a)$ and encourage you to prove $b)$ as an exercise. Let $x,y\in A$ such that $f(x) = f(y)$. Then $(g\circ f)(x) = (g\circ f)(y)$ which implies $x = y$ since $g\circ f$ 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 $f\colon A\to B$ is invertible if and only if it is bijective.

Before the proof, let's see how this answers your question. Assume that $f\colon A\to B$ is invertible, i.e. there exists $g\colon B\to A$ such that $g\circ f=\operatorname{id}_A$, $f\circ g = \operatorname{id}_B$. 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 $b\in B$ there is unique $a\in A$ such that $f(a) = b$. Define $g\colon B\to A$ with $g(b) = a$. Then for any $a\in A$, $g(f(a)) = a$ by definition of $g$ and we have $f(g(b)) = f(a) = b$, for all $b\in B$, i.e. $g\circ f=\operatorname{id}_A$ and $f\circ g = \operatorname{id}_B$. Hence, $f$ is invertible with $g$ its inverse.

For the other direction, use the Lemma.