Is this proof using Cantor's Diagonal Argument correct?

Question

We use $\sim$ to indicate to sets being bijective to eachother, i.e. having the same cardinality in this context. There exists $\psi:\mathbb{N}^2\mapsto\mathbb{Q}$ ,via $\left(a,b\right)\mapsto\frac{a}{b}$ for $b\geq 1$. (For $b=0$, we set $\psi$ as $1$.) Now: $\mathbb{N}\sim \mathbb{N}^2$, hence there exists surjective $\psi\ast:\mathbb{N}\mapsto \mathbb{Q}$. Further,there exists: $\xi:\mathbb{Q}\mapsto \mathbb{N}$, surjective; since $\mathbb{N}\subset \mathbb{Q}$. By the existence of $\psi\ast$ we have: $\mathbb{N}\sim \mathbb{N}^2\preceq \mathbb{Q}$; by virtue of existence of $\xi$: $\mathbb{N}\preceq \mathbb{Q}$, hence also $\mathbb{N}\sim \mathbb{Q}$. I came up with this draft of a proof, but I am not sure, if I used all symbols correctly, since I am only starting to hear this lecture. The prof used a different enumerative argument, yet this one seemed convincing to me as well. Does ist hold true in general that if with two sets $A$ and $B$ are have surjections $\alpha:A\mapsto B$, $\beta:B\mapsto A$ that the two sets are bijective? This might be trivial, yet I was unsure, leading me to pose this question.

Answer 1

Yes, this is true, but it requires the axiom of choice. Basically you want to invoke the Cantor-Bernstein theorem that if $i:A→B$ and $j:B→A$ are injections, then there exists a bijection $\varphi:A→B$. The axiom of choice tells you that you have a surjection $α:A→B$ if and only if there exists an injection $j:B→A$.