Is the empty function always a bijection? Let $f_A:\emptyset\to A$ be the empty function with range $A$. The definition of a bijection as applied to this function is:
$$\forall x,y \in \emptyset (x=y \implies f_A(x)=f_A(y))$$
negating you get:
$$\exists x,y \in \emptyset (x = y\land f_A(x) \neq f_A(y))$$
Which is obviously a false statement since there are no elements in $\emptyset$ at all.
I got troubled by this question when considering the empty set as an inital object of the category Set and the following theorem:
"if I is an initial object then any object isomorphic to I is also an initial object."
but since every empty function is a bijection and thus an isomorphism it follows that all the objects in Set are initial which is obviously false.
What did i miss? 
 A: Keenan Kidwell said it all.
But since you gave a definition of a function on the empty domain (rather than that of a bijection) I'd like to list why every empty function is an injection and why the empty function is a surjection iff the range is empty:
$f:X \rightarrow Y$ is one-to-one iff $ (\forall x_1, x_2 \in X) \: x_1 \neq x_2 \rightarrow f(x_1) \neq f(x_2).$ 
If $X$ is empty the statement is vacuously true because there are no $x_1, x_2 \in X$.
$f:X \rightarrow Y$ is onto iff $(\forall y \in Y)(\exists x \in X) \: y = f(x).$ 
If $X$ is empty and $Y$ is not there is no $x \in X$ such that whatever statement holds. 
If they're both empty the statement is vacuously true, since there are no $y \in Y.$
A: Your symbolic definition of bijectivity is incorrect. The condition you wrote holds because $f_A$ is a function (vacuously in the case of the empty set as domain). A function $f:A\rightarrow B$ is injective (resp. surjective) if and only if $f(x)=f(y)$ implies $x=y$ (resp. for each $z\in B$ there exists $x\in A$ with $f(x)=a$). The unique function from the empty set to any other set is injective, but can be surjective if and only if the target is empty as well. 
