Prove: If $f$ and $g$ are bijective, then $g\circ f$ is bijective.

Question

I want to prove this theorem:

If $f: X \to Y$ and $g: Y \to Z$ are both bijective, then $g \circ f: X \to Z$ is bijective, as well.

Up until now, I've only used induction for proofs, but that isn't applicable here. Any ideas on how to proceed?

Answer 1

Proof. Assume that $f:X\to Y$ and $g:Y\to Z$ are bijective. We Show that $g\circ f$ is bijective by showing that $g\circ f$ is injective and then surjective.

Let $x_1,x_2\in X$ and assume that $(g\circ f)(x_1) = (g\circ f)(x_2)$ equivalently we have $g(f(x_1)) = g(f(x_2))$ this together with the fact that $g$ is bijective and therefore injective it follows that $f(x_1) = f(x_2)$ and via the same considerations about $f$ we may deduce that $x_1 = x_2$.

Now Let $z\in Z$ the surjectivity of $g$ implies that for some $y\in Y$ we have $g(y) = z$ and by the same consideration concerning $f$ it follows that for some $x\in X$ we have $f(x) = y$ consequently $g(f(x)) = z$.

$\blacksquare$

Answer 2

Hint. Show that $g \circ f$ is injective and surjective by using the fact that both $f$ and $g$ are injective and surjective.

Injective. Let $x_1,x_2\in X$ and assume that $g(f(x_1))=g( f(x_2))$. Prove that $x_1=x_2$.

Surjective. Let $z\in Z$, then prove that there is $x\in X$ such that $g (f(x))=z$.

Answer 3

If $f, g$ are both bijective, then there exist inverses $f^{-1} : Y \to X$ and $g^{-1} : Z \to Y$.

Then we have $f^{-1} \circ g^{-1}$ is an inverse to $g \circ f$ because they compose to the identities of $X$ and $Z$:

$$ (f^{-1} \circ g^{-1}) \circ (g \circ f)(x) = (f^{-1} \circ f)(x) = x $$ and $$ (g \circ f) \circ (f^{-1} \circ g^{-1})(z) = (g \circ g^{-1})(z) =z$$ So $g \circ f$ is a bjiection.