Question

A and B are subsets of R where f:A-->B is a function.Prove that f is a bijection if and only if f^-1:B-->A(f inverse from B to A) exists as a bijection

Answer 1

ANSWER on Question #74168 – Math – Functional Analysis

$A$ and $B$ are subsets of $\mathbb{R}$ where $f: A \to B$ is a function. Prove that $f$ is a bijection if and only if $f^{-1}: B \to A$ ($f$ inverse from $B$ to $A$) exists as a bijection.

SOLUTION

Let us recall several necessary definitions

**Definition 1.** Let $f: A \to B$. We say that $f$ is surjective if for all $b \in B$, there exists an $a \in A$ such that $f(a) = b$.

**Definition 2.** Let $f: A \to B$. We say that $f$ is injective if whenever $f(a_1) = f(a_2)$ for some $a_1, a_2 \in A$, then $a_1 = a_2$.

**Definition 3.** We say that $f$ is bijective if it is both injective and surjective.

**Definition 4.** Let $f: A \to B$. A function $f^{-1}: B \to A$ is the inverse of $f$ if

$$f \circ f^{-1} = 1 \text{ and } f^{-1} \circ f = 1.$$

Let us prove two theorems:

**Theorem 1.** Let $f: A \to B$ be bijective. Then $f$ has an inverse.

**Proof.** Let $f: A \to B$ be bijective. We will define a function $f^{-1}: B \to A$ as follows.

Let $b \in B$. Since $f$ is surjective, there exists $a \in A$ such that $f(a) = b$. Let $f^{-1}(b) = a$. Since $f$ is injective, this $a$ is unique, so $f^{-1}$ is well-defined.

Now we much check that $f^{-1}$ is the inverse of $f$.

First we will show that $f^{-1} \circ f = 1$. Let $a \in A$. Let $b = f(a)$.

Then, by definition, $f^{-1}(b) = a$. Then $f^{-1} \circ f(a) = f^{-1}\big(f(a)\big) = f^{-1}(b) = a$.

Now we will show that $f \circ f^{-1} = 1$. Let $b \in B$. Let $a = f^{-1}(b)$.

Then, by definition, $f(a) = b$. Then $f \circ f^{-1}(b) = f\big(f^{-1}(b)\big) = f(a) = b$.

Conclusion,

$$f^{-1} \text{ inverse function to the function } f$$

Theorem 2. Let $f: A \to B$ have an inverse. Then $f$ is bijective.

Proof. Let $f: A \to B$ have an inverse $f^{-1}: B \to A$.

First, we will show that $f$ is surjective. Suppose $b \in B$. Let $a = f^{-1}(b)$.

Then $f(a) = f\big(f^{-1}(b)\big) = f \circ f^{-1}(b) = 1 \cdot (b) = b$. So $f$ is surjective.

Now, we will show that $f$ is injective. Let $a_1, a_2 \in A$ be such that $f(a_1) = f(a_2)$.

We will show $a_1 = a_2$. Let $b = f(a_1)$. Let $a = f^{-1}(b)$.

Then,

$$a_2 = 1 \cdot (a_2) = f^{-1} \circ f(a_2) = f^{-1}(f(a_2)) = f^{-1}(b) = a.$$

But at the same time,

$$a_1 = 1 \cdot (a_1) = f^{-1} \circ f(a_1) = f^{-1}(f(a_1)) = f^{-1}(f(a_2)) = f^{-1}(b) = a.$$

Therefore, $a_1 = a_2$ and we have shown that $f$ is injective.

Conclusion,

We have proved that if a function has the inverse, then it is a bijection.

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