Prove that the functor $V : Cat → Set$ that sends a category $\mathcal{C}$ to its set of arrows, is represented by the category ${0 \rightarrow 1}$ with two objects and a single non identity morphism between them; do not forget to prove that the bijection $V(\mathcal{C}) \cong Hom_{Cat}({0 \rightarrow 1}, \mathcal{C})$ you found is natural!"
Prove that there exists a category of relations, where the objects are all sets, and given sets $A, B$, the set of morphisms $A \rightarrow B$ is the powerset of $A \times B$. Given a relation $R \in \mathcal{P}(A \times B)$ and a relation $S \in \mathcal{P}(B \times C)$, define the composition $S \circ R \in \mathcal{P}(A \times C)$ as the subset $\{(a,c)\in A \times C \ | \ \exists \ b \in B$ such that $(a,b)\in R, (b,c)\in S \}$. Check that this defines an associative composition operation, i.e. $(S \circ R) \circ T = S \circ (R \circ T)$ whenever it makes sense. Given this composition law, there is only one possible choice for what is the identity relation $I \subseteq A \times A$: which one?
These are the types of questions I am tying to solve. I start by listing all the facts I know, the lemmas I can use, the properties of natural transformations, functors etc.
So for the 1st question, $V$ sends a category to its set of morphisms , and we want to prove that $V(\mathcal{C}) \cong Hom_{Cat}(0 \rightarrow 1, \mathcal{C} )$, the $Hom$-functor that maps $\mathcal{C}$ to the set of morphisms between $0 \rightarrow 1$ and $\mathcal{C}$. In $0 \rightarrow 1$, we have only two morphisms, identity. I know that we need to prove the square from $V(\mathcal{C})$ to $Hom(0 \rightarrow 1,\mathcal{C})$, to $V(\mathcal{D})$ to $Hom(0 \rightarrow 1,\mathcal{D})$ commutes, that the natural transformation is an isomorphism(has a well defined inverse)
$$ \require{AMScd} \begin{CD} V(\mathcal{C}) @>a^{\mathcal{C}} >> Hom(0 \rightarrow 1, \mathcal{C})\\ @V {V(F)} VV @VV {Hom(0 \rightarrow 1,F)} V \\ V(\mathcal{D}) @>a^{\mathcal{D}} >> Hom(0 \rightarrow 1, \mathcal{D})\\ \end{CD} $$
And after that, I am totally lost.
I will gladly accept any general tips on approaching these types of questions as well as more specific tips, especially for the 2nd question
It looks to me like you have some misunderstandigs about the definitions. So maybe it will help you when I describe the components of the natural transformation explicitly, you can then try to check that it is indeed natural.
I will denote the category $a \to b$ with two objects and one non-identity arrow by $\mathbf 2$. The category has exactly two objects $a$ and $b$, two identity arrows $id_a$ and $id_b$ and a single non-identity arrow $a\to b$, I give it the name $f$. The composition operations in $\mathbf 2$ are forced, for example $f\circ id_a$ must be $f$ because $\mathbf 2$ is a category. You sometimes write $\{0,1\}$ or even $(0,1)$ for that category, which looks a bit weird to me.
What data does a functor $F:\mathbf 2\to \mathcal C$ contain? The functor $F$ must send the arrow $f$ to some arrow $F(f)$ in $\mathcal C$. The action on the remaining objects and arrows is forced. We must have that $F(a)$ and $F(b)$ must be the domain and codomain of $F(f)$ and that $F(id_a)$ and $F(id_b)$ are the identities of domain and codomain. Let me write $\mathcal C_1$ for the set of arrows in $\mathcal C$, i.e. $\mathcal C_1 = V(\mathcal C)$ in your notation. We see that there is a bijection $$\mathcal C_1 = Cat(\mathbf 2,\mathcal C)$$ Given an arrow $g:x\to y$ in $\mathcal C$, we can define a functor $F:\mathbf 2\to \mathcal C$ by setting $F(a) = x, F(b)=y,F(f) = g$, $F(id_a) =id_x$ and $F(id_b) = id_y$. Conversely, given a functor $F:\mathbf 2\to \mathcal C$ we get an arrow by evaluating $F$ at $f$. The arrow is $F(f) : F(a) \to F(b)$. You can check that the two functions are inverse to each other. I have described the component of the natural isomorphism at $\mathcal C$ for you, your job is now to check that it is natural. This means that given a functor $G:\mathcal C \to \mathcal D$, you have to check that the corresponding square of functions commutes. Remember that $Cat(\mathbf 2,G)$ acts by post-composition on functors. I hope this is enough to get you started. If not, write me a comment! :)