Universal Property of Natural Transformations Proof Verification/ Proof tips Let $\phi$ be a natural transformation between functors $F, F':\mathscr{C} \to \mathscr{D}$, $\tau :Arr\mathscr{D} \to \mathscr{D}$ be the identity natural transformation between the objects of the category of arrows of $\mathscr{D}$ to the morphisms of $\mathscr{D}$ by $\tau(\alpha)=\alpha$, and $\Phi :\mathscr{C} \to Arr\mathscr{D}$ be a functor which sends $c\in \mathscr{C}$ to $\phi(c):Fc \to F'c$
I am asked to show that $\phi=\tau \Phi$, and that for any other functor $\Psi:\mathscr{C} \to Arr\mathscr{D}$ which satisfies $\phi=\tau \Psi$, we must have $\Psi=\Phi$.
I am only asking for the verification of my attempt at the second portion of the proof because the first part is obvious.
Proof: Suppose BWOC there exists a $\Psi \neq \Phi$ such that $\Psi:\mathscr{C} \to Arr\mathscr{D}$ and $\phi=\tau \Psi$. Then for any $c\in \mathscr{C}$ we must have $\tau \Psi(c)=\phi(c)=\tau \Phi(c)$. Since $\tau$ is the identity natural transformation on objects of $Arr\mathscr{D}$, we must have that $\Psi=\Phi$. 
Is there anything else that needs to be shown?
I still feel slightly uncomfortable with these kinds of proofs and any advice on attacking them would be much appreciated.
Thanks.
 A: The statement basically says that $\tau:\text{dom}\Rightarrow\text{cod}:\text{Arr}\mathscr D\to\mathscr D$ is a universal natural transformation between functors to $\mathscr D$, that is, given any natural transformation $\phi:F\Rightarrow F':\mathscr C\to\mathscr D$, there is a unique functor $\Phi:\mathscr C\to\text{Arr}\mathscr D$ such that $\phi=\tau\Phi$
If $\Phi$ is such a functor, then it must send an object $c$ to an object of $\text{Arr}\mathscr C$ at which $\tau$'s component is $\phi_c$, hence $\Phi(c)=\phi_c$. On an arrow $f:c\to c'$ we want $\Phi(f)$ to be a pair of arrows $Fc\to Fc'$ and $F'c\to F'c'$ whose image under $\text{dom}$ resp. $\text{cod}$ is $Ff$ resp. $F'f$, so we can say that $\Phi$, if it exists, must send an arrow $f:c\to c'$ to the commutative square
$$\begin{matrix}
Fc & \xrightarrow{Ff} & Fc' \\
\hspace{10pt} \downarrow\phi_c & & \hspace{10pt}\downarrow\phi_{c'} \\
F'c & \xrightarrow{F'f} & F'c'
\end{matrix}$$
But this obviously defines a functor such that $\tau\Phi=\phi$.
It is important to keep in mind that a natural transformation $\phi$ is not uniquely determined by its components. In fact, two transformations $\sigma$ and $\phi$ between functors from $\mathscr C$ to $\mathscr D$ may have exactly the same components while still being different because they go between different functors (with the same object functions, though). This is reminiscent to a function not being determined solely by a rule $x\mapsto f(x)$, you also need to specify what the domain and codomain are. So instead of just $f:x\mapsto x^2$ you should write $f:\Bbb R\to\Bbb R:x\mapsto x^2$, for example. This becomes evident when we see a natural transformation not just as an $\text{ob}\mathscr C$-indexed family of arrows, but as a morphism in the category of functors from $\mathscr C$ to $\mathscr D$, so this transformation comes together with source and target functors.
So altogether, what you forgot to check was that $\Phi$ and $\Psi$ are equal also on the arrows.
