Does there exist a bijection between empty sets? 
Does there exist a bijection between empty sets?

What I think is:
Since 'for every $x\in \emptyset$, there exists $y\in \emptyset$ such that $(x,y)\in f$' is false and the negation of this statement is also false because of 'existence' sentence..  
But I think empty sets are equipotent intuitively.
Am I wrong? Help
 A: Note that there are not "empty sets", because the axiom of extensionality tells us that all empty sets are equal, as they have the same elements.
Now it is obvious that there is a bijection of a set to itself. The identity, furthermore if you think of it a little bit you will see that the empty set itself is a function whose domain and range are empty, and therefore it is a bijection between that set and itself.
Further reading:


*

*Why is an empty function considered a function?
A: If a statement starts with '$\forall x \in \emptyset$,...', it is vacuously true. 
Because the set contains no elements, any statement holds for 'all elements' in it.
I'm not sure if this answer is 'mathematical' enough, but I think that's the point you might have missed.
A: Yes, there is.
(As said in the comments) the function $f=\emptyset$ is the bijection you seek.
It may seem strange that $f$ is a set, but this is ok since the definition of a function is a set of pairs (s.t ...), in this case we take $f$ to be an empty set.
The definition of $f$ is fine since there are no $x, y_1,y_2 \in \emptyset$ s.t $y_1\neq y_2$ and s.t $(x,y_1),(x,y_2)\in f$.
$f$ is injective since there are no two different elements in $\emptyset$ that $f$ maps to the the same element (this is just because there are no two different elements in $\emptyset$).
$f$ is surjective since for each $y$ in  $\emptyset$ there exist a source (this is since there are no elements in $\emptyset$ hence we can't say there exist an element that doesn't have a source). 
