Your question sounds like your preferred foundation is set theory, so let me speak in terms of set theory. A map $f : A \to B$ between sets is a functional relation, i.e., a subset $f \subseteq A \times B$ satisfying:
1. Totality: $\forall x \in A . \exists y \in B . (x,y) \in f$
2. Single-valuedness: $\forall x \in A . \forall y, z \in B . ((x,y) \in f \land (x,z) \in f \implies y = z)$.
We usually write $f(x) = y$ instead of $(x,y) \in f$.
The same definition applies to classes. A map $F : C \to B$ between classes $C$ and $D$ is a subclass of $C \times D$ which is total and single-valued.
Exercise (allowed since this is not a research question): the domain and codomain of a function $F : C \to D$ cannot be recovered from the functional relation $F : C \to D$. F$. (If$C$and$F$are empty, how do we recover$D$?) Therefore, the object part of a functor must be a triple$(C,D,F)$rather than just$F$. But how can we form ordered triples of classes? 1 Your question sounds like your preferred foundation is set theory, so let me speak in terms of set theory. A map$f : A \to B$between sets is a functional relation, i.e., a subset$f \subseteq A \times B$satisfying: 1. Totality:$\forall x \in A . \exists y \in B . (x,y) \in f$2. Single-valuedness:$\forall x \in A . \forall y, z \in B . ((x,y) \in f \land (x,z) \in f \implies y = z)$. We usually write$f(x) = y$instead of$(x,y) \in f$. The same definition applies to classes. A map$F : C \to B$between classes$C$and$D$is a subclass of$C \times D$which is total and single-valued. Exercise (allowed since this is not a research question): the domain and codomain of a function$F : C \to D$cannot be recovered from the functional relation$F : C \to D$. (If$C$and$F$are empty, how do we recover$D$?) Therefore, the object part of a functor must be a triple$(C,D,F)$rather than just$F\$. But how can we form ordered triples of classes?