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.
 A: Let $f:A\to\varnothing$ denote a function. 
If $a\in A$ then automatically $f(a)\in\varnothing$. This contradiction makes us conclude that $A=\varnothing$.
There is indeed a map $f:\varnothing\to\varnothing$. Its graph is a subset of $\varnothing\times\varnothing=\varnothing$ hence is $\varnothing$ itself. This shows that $f$ is unique. It is called the empty map. This map is vacuously bijective.

edit:
Every function $f:A\to\varnothing$  is a bijection.
Let $f:A\to\varnothing$ be a function. Then for every $z\in\varnothing$ there is a unique $x\in A$ such that $f(x)=z$. 
Formally:$$\forall z\;[z\in\varnothing\implies\exists!x[x\in A\wedge f(x)=z]]$$
This statement is vacuously true. This because for every $z$ it is false that $z\in\varnothing$. We cannot find any $z\in\varnothing$ for which it is not true (even stronger: we cannot find any $z\in\varnothing$ at all). And the statement expresses that $f$ is injective and surjective as well.
A: A function $A\longrightarrow\emptyset$ is trivially a relation. i.e. a subset of $A\times\emptyset =\emptyset$. Then, the only possible relation is $\emptyset$, but $\emptyset$ isn't a function except if $A = \emptyset$ (why? hint: what is $f(a)$ for some $a\in A$?).
A: The fallacy is there exists no function $f:A\rightarrow\varnothing$ (A function from $A$ to $B$ is a subset of $A\times B$ such that for any $a\in A$ there exists a unique $b\in B$ such that $(a,b)$ is in the subset) if $A$ is not empty.
A: 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 $A=\emptyset$.
A: We will prove that every function $f:A\rightarrow \emptyset$ is a bijection. Observe that there are 2 cases: either $A\neq \emptyset$ or $A=\emptyset$.
Case 1: We will show that if $A\neq \emptyset$ then there does not exist a function $f:A\rightarrow \emptyset $, and so all 0 of these functions are vacuously bijections. 
By way of contradiction, assume that $A\neq \emptyset$ and $\exists f:A\rightarrow \emptyset $.  Since $A$ is not empty, $\exists a\in A$, and since the codomain is the empty set, we have that $f(a) \in \emptyset $. Since the empty set has no elements, this is a contradiction. Therefore the claim is proved. $\square$
If you want to use the more rigorous definition of a function, you could write this proof as follows:
Suppose by way of contradiction that $A\neq\emptyset$
 , and $f:A\rightarrow\emptyset$
 . Recall that a function $f:A\rightarrow B$
  is a relation and is thus a subset of the cartesian product $A\times B=\{(a,b):a\in A\land b\in B\}$
 . Since $B=\emptyset$
 , $\exists b\in B$
  is false. However, by the definition of a function, $\forall a\in A,\exists b\in B(a,b)\in f$
 , which is a contradiction, since $\neg\exists b\in B$
 . Thus the claim is proved. $\square$
Thus, by either proof, we can see there are no functions $f:A\rightarrow \emptyset $ for $A\neq \emptyset$. Thus we have (vacuously) that all ($0$) of these functions are bijections, and unicorns, and superheroes, etc. 

Case 2: Let $A = \emptyset$ and $f:A\rightarrow \emptyset$. We will show that $f=\emptyset$
  is a bijection from $\emptyset\rightarrow\emptyset$
To prove this claim, we need to use the rigorous definition of a function $f:A\rightarrow B$ as a type of relation, ie a subset of the Cartesian product $A\times B$, in which the following are true
1) $\forall a\in A,\exists b\in B((a,b)\in f)$
2) [$(a,b)\in f$ and $(a,b')\in f] \implies b=b'$
Since there are no $a\in A=\emptyset$
 , the first condition is vacuously true. Likewise, since $f=\emptyset$
 , the antecdent/hypothesis $(a,b)\in f$
  and $(a,b')\in f$
  is false, and therefore the second condition is vacuously true, and thus $f=\emptyset$
  is a function.
Since there are no elements in either the domain or the codomain, we get bijectivity vacuously as the OP correctly noted. $\square$
