Tao Analysis I - 3.3.2

Exercise 3.3.2

Let $f: X \mapsto Y$ and $g: Y \mapsto Z$ be functions. Show that if $f$ and $g$ are both injective, then so is $g \circ f$; similarly, show that if $f$ and $g$ are both surjective, then so is $g \circ f$.

Injective $g \circ f$

By Definition 3.3.17, a function $f$ is injective, one-to-one, if

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

Let's consider $g \circ f$ being injective,

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

Since $g$ is injective, $g(y)=g(y') \implies y=y'$, which here means

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

Now, since $f$ is injective, $f(x) = f(x') \implies x=x'$.

Thus, we have shown 

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

which means $g \circ f$ is injective. $\square$

Surjective $g \circ f$

By Definition 3.3.20, a function $f: X \mapsto Y$ is surjective, onto, if

$$(\forall y\in Y)(\exists x \in X)(f(x)=y)$$

In other words, every element of the codomain is mapped to by an element in the domain.

Let's consider $g \circ f$ being surjective. We need to show that every element $z \in Z$ has an element in $x \in X$ such that $g(f(x))=z$.

• Since $g$ is surjective, we know that every element $z \in Z$ has an element $y \in Y$ such that $g(y) = z$.
• Since $f$ is surjective, we know that every element $y \in Y$ has an element $x \in X$ such that $f(x)=y$.

We can use the second statement to rewrite the first as: every element $z \in Z$ has an element $x \in X$ such that $g(f(x)) = z$.

That is, $g \circ f$ is surjective. $\square$