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.
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 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 $B$.
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 $f(a) = b$.
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.
