If $A, B, C, D$ are sets such that $A \sim B$ and $C \sim D$, $\exists$ bijections $f: A \to B$ and $g: C \to D$. Let $h: A \times C \to B \times D$ be $h(a,c) = (f(a), g(c))$. Show that $h$ is a bijection (and thus $A\times C \sim B \times D$).
How can I solve this?
1) $h$ is injective: $h(a,c) = h(a', c') \implies \big(f(a),g(c)\big) = \big(f(a'),g(c'\big)$, so $f(a) = f(a')$ and $g(c) = g(c')$. Since $f,g$ are bijective, they must be injective, so $a=a'$ and $c=c'$, therefore $h$ is injective.
2) $h$ is surjective: since $f,g$ are bijective, they must be surjective, so for arbitrary $b \in B, d \in D$ there exist $a \in A, c \in C$ such that $f(a) = b, g(c) = d$, but this means preciselz that $h(a,c) = (b,d)$, so $h$ is surjective.
Being injective and surjective, $h$ must be bijective.
We know that if a function has inverse, then it is a bijection. See:
So it suffices to show that $h$ has inverse. In this case, it is very easy to guess what the inverse looks like. Let us define $h' \colon B\times D \to A\times C$ by $$h'(b,d)=(f^{-1}(b),g^{-1}(d)).$$ (Since $f$, $g$ are bijections, we know that $f^{-1}$, $g^{-1}$ exist.)
So it remains to verify that $h'$ is inverse to $h$; i.e. that both $h'\circ h$ and $h\circ h'$ are equal to identity. If you show this (it should be pretty straightforward) you are done.
