It is posible to have a function $f$ with the following property? I am considering the function $f$ defined as
$$f:\emptyset\to\emptyset$$
My thought is that a function maps elements from a set to another one, but the empty set has no elements to map, so I think there cannot exist a function $f$ that has this property, but I don't know how to prove it and I'd like some light on this.
There is a function $I$ though which can map the empty set to the empty set if its domain and codomain are the set containing the empty set. But it is not the same question.
 A: There is a unique function $\emptyset \to \emptyset$. 
The reason is that formally a function $f: A \to B$ is a subset of $A \times B$ such that for every $a \in A$ there is exactly one pair $(a,b) \in f$; normally one writes $b = f(a)$. But if $A$ is empty, then $A \times B$ is also empty, so we can (and only can) take as our empty function the empty set $\emptyset \subseteq \emptyset \times B = \emptyset$.
See here for further details. 
A: A function from $A$ to $B$ is a subset $f$ of $A\times B$ such that, for each $a\in A$, there is one and only one $b\in B$ such that $(a,b)\in f$. Therefore, if $A=\emptyset$ then, no matter what $B$ is, $\emptyset$ is a function from $A$ to $B$.
A: 
Definition: A function $f$ mapping from a set $A$ to a set $B$ is a subset of $A \times B$ such that for all $a$ in $A$, there exists an element $b$ in $B$ with $(a,b) \in f$ and if there exists an element $b^\prime$ in $B$ with $(a,b^\prime) \in f$, then $b = b^\prime$ (i.e., every element in $A$ is uniquely mapped to an element in $B$).

Now, since $\emptyset \times B = \emptyset$, any function $f$ mapping from $\emptyset$ to $B$ must be $f = \emptyset$.

Note: Of course, the notation $(a,b) \in f$ is quite cumbersome, so we write instead $f(a) = b$.

A: While this is an illustration for some non-empty function from some set $X$ to $Y$..

.. for the empty function from $X=\emptyset$ to the same set $Y$ you would have

It does not matter what set $Y$ is, the relevant fact is that the function graph set (depicted by the set of arrows in the images) is empty.
