Graphs from empty functions Let us assume that
\begin{align*}
f \colon \emptyset \to \emptyset\,,\quad g \colon \emptyset \to A\quad\text{and} \quad h \colon A \to \emptyset\,,
\end{align*}
where $A$ is a non-empty set.
I am interested in the graphs of the functions, i.e. the sets $f$, $g$ and $h$. IMO it holds that $f = g = h = \emptyset$. Am I right and if not then why?
 A: You are correct.
The graph of a function $\psi:X\to Y$ is by definition a (particular) subset of $X\times Y$. If either $X$ or $Y$ is $\varnothing$, then the entire $X\times Y$ is $\varnothing$, so the graph of $\psi$ must be $\varnothing$ also.
A: When working with binary relations that include the empty set, special care is needed when you are identifying which binary relations are functions. Let us go by parts.
First, recall the definition of a binary relation.
Definition. Let $A$ and $B$ be any sets. A binary relation from $A$ to $B$ is any subset of the set $A \times B$.
So, the binary relations from a set $A$ to a set $B$ are exactly the elements of the set $\mathcal{P}(A \times B)$ (i.e., the power set of $A \times B$). A function from a set $A$ to a set $B$ is a special case of binary relations. The definitions goes as follows.
Definition. Let $A$ and $B$ be any sets and let $R \subseteq A \times B$ be a binary relation from $A$ to $B$. We say that $R$ is a function from $A$ to $B$ if

*

*$D_{R} = A$, where $D_R = \{x \in A \mid \exists y \in B \colon xRy\}$ is the domain of $R$;


*$xRy \wedge xRz \implies y=z$.
It is straightforward to note that $R$ is a function if $\forall x \in A, \exists^1 y \in B \colon xRy$.
Now, let’s see what we have in your question. Let $A$ be a non-empty set and let $F \subseteq \emptyset \times \emptyset$, $G \subseteq \emptyset \times A$ and $H \subseteq A \times \emptyset$ be binary relations. Let us look at each of these binary relations.
For the binary relations $F$ and $G$, the propositions $\forall x \in \emptyset, \exists^1 y \in \emptyset \colon xFy$ and $\forall x \in \emptyset, \exists^1 y \in A \colon xGy$ are vacuously true. Note that “$\forall x \in \emptyset$“ is an impossible condition. So the antecedent of each of those propositions are false, which makes the whole proposition true. (This is logic, any doubt on this, you can ask that I will gladly answer).
Now, for the relation $H$ the situation is different. Let us look at the proposition
$$\forall x \in A, \exists^1 y \in \emptyset \colon xHy.$$
This is not a function according to the definition. Since $A \neq \emptyset$, then there exists some object $a \in A$. Although, since the $\emptyset$ is devoid elements, there will not be an element, say $b$, in $\emptyset$ such that $aHb$. Therefore, the proposition is false, which means that $H$ is not a function.
Although, by definition of cartesian product, we have that $F = G = H = \emptyset$, only $F$ and $G$ are functions.
To think about. If you denote the set of functions from a set $X$ to a set $Y$ by $Y^X$, you can easily check that $Y^{\emptyset} = \{\emptyset\}$, for any set $Y$, and $\emptyset^{X} = \emptyset$, for any non-empty set $X$. In particular, we have that $\emptyset^{\emptyset} = \{\emptyset\}$. Try this. If you have any doubt, you can ask.
