If $A \sim B$ and $B \sim C$. Prove that $A \sim C$
What I have:
We know there exists functions $f,g$ such that $f:A\to B$ and $g:B \to C$ where $f$ and $g$ are bijective.
We thus require to show that there exists a function $h:A \to C$ where $h$ is bijective.
Can anyone please give me some hints that might point me in the right direction? I do not want a complete answer, simply a nudge in the right direction.
UPDATE: This is my (very rough) attempt at the proof.
We know there exists functions $f,g$ such that $f:A\to B$ and $g:B \to C$ where $f$ and $g$ are bijective.
We thus require to show that there exists a function $h:A \to C$ where $h$ is bijective.
Assume that $h(x) = g \circ f = g(f(x))$.
We must now show that $h$ is bijective.
$\underline{RTP:}$ If $f$ and $g$ are surjective, then $g \circ f$ is surjective.
$\underline{Proof:}$ We are given that $f$ and $g$ are surjective (since they are bijective). Suppose $c \in C$, then since $g$ is surjective, we know $\exists b \in B$ such that $g(b) = c$. Similarly, since $f$ is surjective, $\exists a \in A$ such that $f(a) = b$.
Thus \begin{align}g \circ f(a) &= g(f(a)) \\ &= g(b) \\ &= c\end{align} Hence we have shown that $h$ is surjective.
We must now show that $h$ is injective.
$\underline{RTP:}$ If $f$ snd $g$ are injective, then $g\circ f$ is also injective.
$\underline{Proof:}$ We are given that $f$ and $g$ are injective (since they are bijective) and suppose $$g \circ f(x_1) = g \circ f(x_2)$$ We thus have that $$g(f(x_1))= g(f(x_2))$$ Since $g$ is injective, we know that this is true if and only if $f(x_1) = f(x_2)$. Similarly, since $f$ is injective, it follows that $$x_1 = x_2$$ Thus $g \circ f$ is injective.
We have thus proven that there exists a function $h(x)= g \circ f : A \to C$ that is bijective.
Hence we can conclude that $A \sim C$.