Evaluating correctness of various definitions countable sets I was trying to understand the definition of countable set (again!!!). Wikipedia has a very great explanation:


  
  A set $S$ is countable if there exists an $\color{red}{\text{injective}}$ function $f$ from $S$ to the natural numbers $\mathbb N$.
  If such an $f$ can be found that is also surjective (and therefore $\color{red}{\text{bijective}}$), then $S$ is called countably infinite.
  In other words, a set is countably infinite if it has $\color{red}{\text{bijection}}$ with the $\mathbb N$.
  


So I summarize:


$S$ is countable iff $S\xrightarrow{injection}\mathbb N$
$S$ is countably infinite iff $S\xrightarrow{bijection}\mathbb N$


But then wikipedia confuses by stating following points:


  Theorem: Let $S$ be a set. The following statements are equivalent:
  
  
  $S$ is countable, i.e. there exists an injective function $f : S → \mathbb N$. 
  Either $S$ is empty or there exists a surjective function $g : \mathbb N → S$.
  Either $S$ is finite or there exists a bijection $h : \mathbb N → S$.
  


Q1. I feel 2nd statement is wrong, as it allows some element in $S$ to not to map to any element in $\mathbb N$. That is $\mathbb N \xrightarrow{surjection} S$ does not imply $S\xrightarrow{injection}\mathbb N$. Hence $S$ is not countable. Right?
Q2. 3rd  statement defines countably infinite set, so its countable also. Right?
Q3. Also I dont get if the extra restrictions of emptyness and finiteness in statements 2 and 3 are required.

Wikipedia further says:


  Corollary: Let $S$ and $T$ be sets.
  
  
  If the function $f : S → T$ is injective and $T$ is countable then $S$ is countable.
  If the function $g : S → T$ is surjective and $S$ is countable then $T$ is countable.
  


Q4. Here, too, I feel 2nd statement is incorrect for the same reason as 2nd statement in the theorem. Right?
The Wikipedia theorem is correct. More generally, if $A$ is an ordered set and there exists a surjection $g\colon A \twoheadrightarrow B$ then there exists an injection $f\colon B \hookrightarrow A$ given by $f(b) = \min g^{-1}(b)$. This is an injection since $g(f(b)) = b$.

More generally, if you assume the axiom of choice then you don't need to assume that $A$ is ordered, since the axiom of choice guarantees the existence of a function $f\colon B \to A$ satisfying $f(b) \in g^{-1}(b)$ for all $b \in B$.