Tao Analysis I - 3.3.7

Exercise 3.3.7

Let $f : X \mapsto Y$ and $g: Y \mapsto Z$ be functions. Show that if $f$ and $g$ are bijective, then so is $g \circ f$, and we have $(g \circ f)^{−1} = f^{−1}  \circ g^{−1}$.

Show $g \circ f$ is Bijective 

To show $g \circ f$ is bijective, we need to show it is injective and surjective.

For the purpose of contradiction, let's assume $g \circ f$ is not injective. That means there exists $x, x' \in X$, where $x \neq x'$, such that

$$g (f(x)) = g(f(x'))$$

Since $g$ is injective, by Definition 3.3.17, we have 

$$g (f(x)) = g(f(x')) \implies f(x)=f(x')$$

Again, since $f$ is injective, by Definition 3.3.17, we have

$$f(x)=f(x') \implies x=x'$$

This contradicts our assertion that $x \neq x'$.  We have shown that $g \circ f$ is injective.

Again, for the purpose of contradiction, let's assume $g \circ f$ is not surjective. This means there exists a $z' \in Z$ such that

$$g(f(x)) \neq z'$$

We are given that $g$ is surjective, which means every $z \in Z$ has a $y \in Y$ such that $z=g(y)$.  Similarly, we are given that $f$ is surjective, which means that every $y \in Y$ has an $x \in X$ such that $y=f(x)$. 

Together this means every $z \in Z$ has an $x \in X$ such that $z = g(f(x))$. This contradicts our assertion that $g \circ f$ is not surjective. We have shown that $g \circ f$ is surjective.

By showing $g \circ f$ is both injective and surjective, we have shown it is bijective. $\square$

Show $(g \circ f)^{−1} = f^{−1}  \circ g^{−1}$

Let's consider $(g \circ f)^{−1}$. We have just shown it is bijective, so it is invertible.

Using the textbook's definition, $x=(g \circ f)^{−1}(z)$ is the unique value of $x$ such that

$$g(f(x))=z$$

We can apply $g^{-1}$ to both sides,

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

And then simplify using the cancellation law from Exercise 3.3.6,

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

Again, applying $f^{-1}$ to both sides, and using the cancellation law, gives us

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

We have shown $x=(g \circ f)^{−1}(z) = f^{−1}  \circ g^{−1}(z)$, which means $(g \circ f)^{−1} = f^{−1}  \circ g^{−1}$. $\square$