Does there exist a deterministic, invertible function $\{0,1\}^n \rightarrow \{0,1\}^n$ that is not a bijection? One of the requirements for a function to be a PRP is


  
  For any $K \in \{0,1\}^s$, $F$ is a bijection from $\{0,1\}^n \rightarrow \{0,1\}^n$.
  


Taken from Wikipedia

Does this have to be explicitly said? Is it even possible for this not to be true for a deterministic, invertible function where the input and output domains are the same?
Does this have to be explicitly said?


You stripped the preceding context from that quote:


  Let $F$ be a mapping $\{0,1\}^n × \{0,1\}^s → \{0,1\}^n$. $F$ is a PRP if
  
  
  For any $K ∈ \{0,1\}^s$, $F$ is a bijection from $\{0,1\}^n$ to $\{0,1\}^n$.
  [...]
  


Not all mappings $F : \{0,1\}^n × \{0,1\}^s → \{0,1\}^n$ are bijections for all choices of $K$, so yes, the definition certainly has to state the bijectivity requirement.  


  Is it even possible for this not to be true for a deterministic, invertible function where the input and output domains are the same?


You're reading the definition with an implicit assumption in mind that the function must be invertible, but the clause you're pondering is the one that actually states the requirement.