Bijection from $A \rightarrow \varnothing$ My thoughts. We need to prove that:
1 $\forall x,y \in A, \text{ if } f(x) = f(y) \rightarrow x = y$ 
2 $\forall y \in \varnothing, \exists x \in A,  f(x) = y$.  
In (1), $f(x) = f(y)$ is false, since neither $f(x)$ nor $f(y)$ have a value, so (1) is vacuously true. 
Also, $\forall y \in \varnothing, P(x, y)$ is vacuously true. So both statements are true.

Admittedly, Does there exist a bijection between empty sets? offers some guidance, but I am unsure whether my rationale for #1 is sound.

Functions $f:A\rightarrow B$ can be thought of as particular subsets of $A\times B$ (ones that satisfy the well-defined property).  Since $A\times\emptyset=\emptyset$, there is only one subset of $A\times\emptyset$.  
Additionally, for the domain of $f:A\rightarrow B$ to be $A$, for all $a\in A$, there must exist $b\in B$ such that $(a,b)\in f$.  In your case, since $f=\emptyset$, $(a,b)\not\in f$, so it must be that there is no $a\in A$.  Hence $A=\emptyset$.
