Tao Analysis I - 3.3.6

Exercise 3.3.6

Let $f : X \mapsto Y$ be a bijective function, and let $f^{−1} : Y \mapsto X$ be its inverse. Verify the cancellation laws $f^{−1}(f(x))=x$ for all $x∈X$ and $f(f^{−1}(y))=y$ for all $y∈Y$. Conclude that $f^{−1}$ is also invertible and has $f$ as its inverse (thus $( f^{−1})^{−1} = f$).

Let's first remind ourselves of what the inverse of a function is according to the textbook:

If $f$ is bijective, then for every $y \in Y$, there is exactly one $x$ such that $f(x)=y$. This value of $x$ is denoted $f^{-1}(y)$; thus $f^{-1}$ is a function from $Y$ to $X$. We call $f^{-1}$ the inverse of $f$.

Show  $f^{−1}(f(x))=x$

Let's first consider $f(x)$.

Since $f$ is injective, for every $x$ there is a unique $y=f(x) \in Y$. And since $f$ is surjective, for every $y \in Y$, there is an $x$ such that $y=f(x)$. Together, this means every $y$ has a unique $x$ such that $y=f(x)$. We can therefore denote this $x$ as $f^{-1}(y)$, as per the textbook's definition:

$$f^{-1}(y) = x$$

And since $y=f(x)$, we have

$$f^{-1}(f(x)) = x$$

Thus, we have verified the cancellation law $f^{−1}(f(x))=x$. $\square$

Show  $f(f^{−1}(y))=y$

Let's first consider $f(x)=y$.

Since we have shown that $x=f^{-1}(x)$ for all $x \in X$, we can write

$$f(f^{-1}(x)) = y$$

Thus, we have verified the cancellation law $f(f^{−1}(y))=y$. $\square$

Show $f^{−1}$ is Invertible

To show that a function $f$ is invertible, we need to show it is bijective:

• If $f$ is not injective, then the inverse would not be unique. 
• If $f$ is not surjective, then some elements of its codomain $Y$ are not invertible. 

For the purpose of contradiction, let's assume $f^{-1}$ is not injective. That would mean there exist $y, y' \in Y$, where $y \neq y'$, such that

$$f^{-1}(y)=f^{-1}(y')$$

Now, let's use $y=f(x)$ and $y'=f(x')$, to give us

$$f^{-1}(f(x))=f^{-1}(f(x'))$$

We can apply the first cancellation law above,

$$x=x'$$

Because $f$ is given as injective, this means $f(x)=f(x')$, that is $y=y'$. This is a contradiction, because we set $y \neq y'$ earlier.

Thus we have shown $f^{-1}$ is injective.

Now we need to show $f^{-1}$ is surjective.

For the purpose of contradiction, let's assume $f^{-1}$ is not surjective. That means there exists an $x' \in X$ such that

$$f^{-1}(y) \neq x'$$

Applying $f$ to both sides gives us

$$f(f(^{-1}(y)) \neq f(x') = y'$$

Using the second cancellation law gives us

$$y \neq f(x') = y'$$

This is a contradiction because $f$ is surjective, and for every $y \in Y$, including this $y'$,  there exists an $x$ such that $f(x)=y$.

Thus we have shown $f^{-1}$ is surjective.

Having shown $f^{-1}$ is both injective and surjective, that is bijective, we can say it is invertible. $\square$

Show $f$ is the inverse of $f^{-1}$

Let's use the description of an inverse given in the textbook.

If $f^{-1}: Y \mapsto X$ is bijective, then for every $x \in X$, there is exactly one $y$ such that $f^{-1}(y)=x$. This value of $y$ is denoted $(f^{-1})^{-1}(x)$, thus $(f^{-1})^{-1}$ is a function from $X$ to $Y$, and is the inverse of $f^{-1}$.

Let's apply $f$ to both sides of $f^{-1}(y)=x$

$$f(f^{-1}(y))=f(x)$$

Applying the second cancellation law gives,

$$y=f(x)$$

That is, $y$, which is $(f^{-1})^{-1}(x)$, is $f(x)$. That means the inverse of $f^{-1}$ is $f$. $\square$