Composition Of Two Bijections Is a Bijection: Proof

A common exercise from an "Introduction To Real Analysis" course is to show that

1. if $f: A\rightarrow B$ is a bijection, and
2. if $g: B\rightarrow C$ is a bijection

then the composite $g\circ f$ is a bijective map of A onto C.

We just need to show that $g\circ f: A\rightarrow C$ is one to one and onto.

$$g\circ f = g(f(x))$$
$$f(x)=y     (y\in B)$$
$$g(y)=z    (z\in C)$$

First, show the composite is injective (one to one):
by 1, we know that $f(x_{1})=f(x_{2})$ means $x_{1}=x_{2}$ (f is one to one - injective).
by 2, we know that $g(f(x_{1}))=g(f(x_{2}))$ means $f(x_{1})=f(x_{2})$ (g is one to one - injective).

a) So, we see that $g(f(x_{1}))=g(f(x_{2}))$ means that  $x_{1}=x_{2}$, which means that $g\circ f$ is       an injective (one to one) map of A onto C.

Second, show the composite is surjective (onto):
by 1, we know that for any $b\in B$ there exists at least one $x\in A$ such that $f(x)=b$.
by 2, we know that for any $c\in C$ there exists at least one $b\in B$ such that $g(b)=c$.

b) So, we see that for any$c\in C$ there exists at least one  $x\in A$ such that $g(f(x))=c$, which means that    $g\circ f$ is a surjective (onto) map of A onto C.

With the results a) and b) we see that the composite $g\circ f$ is indeed a bijective map of A onto C.

QED