I have some issues with fully understanding the concept of surjection.
For the proof of $A_{m}$ being a countable set for each m $\in$ N, then the union $A = \bigcup_{m=1}^{\infty} A_{m}$ is countable; my textbook states that we let $\beta_{m}$ be a surjection of N onto $A_{m}$ for each m $\in$ N and then define $\gamma$ : N x N -> A , $\gamma(m,n) = \beta_m(n)$. Then, it claims $\gamma$ is a surjection. If we let $a \in A$, then there is a $m \in$ N such that $a \in A_{m}$. Thus, there exists a least $n \in$ N so that $a = \beta_m (n)$. So, it concludes that $a = \gamma(m,n)$ and thus it proves that $\gamma$ is a surjection.
The fact is, why would proving $a = \gamma(m,n)$ make the function surjective? From the definition of a surjective function, if $f$: A -> B is surjective, then $\forall y \in B, \exists x\in A$ such that $f(x) = y$. In this case, how can we prove that for all $a \in A$ this will hold true? For this case, it seems that it is only for a particular $a$.
