I try to verify the following claims from my textbook:
A functor between groups $G$ and $H$ is a group homomorphism from $G$ to $H$.
For the first condition, $F_1f : F_0A \to F_0B$ clearly holds, because there is only one function between the objects of $G$ and $H$ viewed as categories, namely $F_0(G) = H$, therefore $F_1f : H \to H$, which holds because all morphisms of $H$ have the domain and codomain $H$. Secondly, $F_1(f \circ g) = F_1f \circ F_1g$ holds trivially because $F_1$ is a group homomorphism. Thirdly, $F(\mathsf{1}_G) = \mathsf{1}_{F_0G}$ holds because $F_1(\mathsf{1}_G) = F(e_G) = e_H = \mathsf{1}_H = \mathsf{1}_{F_0G}$. The second equality holds due to $F_1$ being a group homomorphism and the fourth because the only element which $F_0$ maps to is $H$.
A functor between posets $P$ and $Q$ is a monotone function from $P$ to $Q$.
A monotone function has the property $x \leq y \Rightarrow f(x) \leq f(y)$. If we view the $\leq$ relation as arrows between objects, the above definition is equivalent to $F_1(A \to B) = F_0A \to F_0B$, which is justified by the fact that in posets as categories, arrows between objects are unique. This implies $F_1f : F_0A \to F_0B$, and the first condition is met. Composition is carried out via transitivity, that is, we can write out $f \circ g$ as $x \leq y \land y \leq z$, which implies (i.e., composes to) $x \leq z$. Hence, $F_1(f \circ g) = F_1(x \leq z)$. On the other hand, $F_1f \circ F_1g$ is what $F_1(x \leq y) \land F_1(y \leq z)$ composes to, which is equivalent to $F_0x \leq F_0y \land F_0y \leq F_0z$, which composes to $F_0x \leq F_0z$, which is equal to $F_1(x \leq z)$, thus proving the third condition.
Perhaps the real challenge of this exercise was to translate the common meaning of "homomorphism" and "monotone map" onto the functor $F$, as it has a richer structure (it is a pair of functions $F_0$ and $F_1$) and hence the translation is not trivial.
Addendum. Since it has been pointed out that my notation is not standard, here is the full definition of a functor I am working with. For a category $\mathcal{C}$, $\mathcal{C}_0$ is the class of objects, and $\mathcal{C}_1$ is the class of morphisms.
