Empty functions are not injective? Many sources say that empty functions such as $f:\emptyset \rightarrow S$ are injective because it is a vacuous truth. But currently I am reading a book on axiomatic set by Patrick Suppes, and he gives a definition of an injective(one-to-to) function that prevents any empty function from being a injective. The definition Suppes gives is
$f$ is one-to-one $\leftrightarrow f$ and $\breve{f}$ are functions
For those that don't know what $\breve{f}$ is, here is the definition
$\breve{f}=\{\langle x,y \rangle: \langle y,x \rangle \in f \}$
According to this definition, any empty function is not injective because $\breve{f}:S \rightarrow \emptyset$ is not a function.
Is Suppes definition correct either though it doesn't allow empty functions to be injective or I'm I missing something?
 A: That's a good question, but you're not correct.
The reason is that if $f\colon\varnothing\to S$, then $f=\varnothing$. Therefore $\breve f=\varnothing$. Therefore $f=\breve f$ and both satisfy the condition for being a function.
It is true, however, that $\breve f$ is not a function whose domain is $S$ (unless $S$ is empty). And note that we shouldn't require it is a function from $S$, because then the function $f(n)=n+1$ as a function from $\Bbb N$ to itself is not injective anymore, which is plain preposterous.
Also, there is but one unique empty function. The empty set.
A: If $f$ is the empty function, then so is $\breve f$.  Its domain, however, is not $S$ but the empty set. So, unless Suppes said that the domain of $\breve f$ is $S$, his definition is correct.  (If he had required $\breve f$ to have domain $S$, then the definition would be wrong not only for the empty function but for any injective function that isn't surjective.)
A: We agree that a function $f:A\to B$ is a subset, $f\subset A\times B$ that satisfies conditions (that we need not repeat). Now let us suppose that $A=\emptyset$. Then the set, $\emptyset\times B$ is itself the empty set since their is no way to make a pair if we cannot put something in the first slot. So $f=\emptyset$. But then we also have $\breve{f}$ must be a subset of $B\times\emptyset$, which must be the empty set. But the empty set is indeed a function from the empty set to any other set.
