# Universal Property of the Tautological Transformation

Let me give some background to the question. For a category $\mathscr{C}$, define the tautological correspondence $$\tau : Ob\space Arr \space\mathscr{C} \longrightarrow Mor\space\mathscr{C}$$ to be the identity correspondance between objects in $Arr\space\mathscr{C}$ and morphisms of $\mathscr{C}$. It can be checked that $\tau$ is a natural transformation between the source and the target functors of $\mathscr{C}$.

The statement to prove is:

If $F$ and $F'$ are functors from $\mathscr{B}$ to $\mathscr{C}$ and $\phi$ is a natural transformation from $F$ to $F'$, then $\phi$ is pulled uniquely from the the tautological correspondance $\tau$ on $Arr\space\mathscr{C}$ defined above. In other words, there exists a unique functor $\Phi:\mathscr{B} \longrightarrow Arr\space\mathscr{C}$ such that $\phi=\tau\circ\Phi$.

I am able to show the existence by constructing such a $\Phi$: assign $\phi_b$ to each object b of $\mathscr{B}$. Assign the pair of morphisms ($F\alpha,F'\alpha$) to each morphism $\alpha:c\rightarrow c'$ of $\mathscr{B}$. Since $\phi$ is a natural transformation the following diagram commute : $\require{AMScd}$ \begin{CD} Fc @>{\phi_c}>> F'c\\ @V{F\alpha}VV @VV{F'\alpha}V\\ Fc' @>>{\phi_{c'}}> F'c' \end{CD}

and it is easy to check that $\Phi$ is a functor from $\mathscr{B}$ to $Arr\space \mathscr{C}$.

Now the part I'm stuck on is uniqueness. Perhaps, I can use the fact that $\tau$ is a natural transformation between the source and the target functors of $\mathscr{C}$. But I don't know how to carry this through. Any help would be appreciated.

As a beginner of Category Theory, I apologize for any sloppiness that would baffle those fluent in the language.

For any $b\in Ob\mathcal B$, we must have $\phi(b)=\tau(\Phi(b))$ but $\tau$ is the identical map of arrows of $\mathcal C$, so $\phi(b)=\Phi(b)$.
For morphisms, we indeed have to interpret $\tau$ as a natural transformation $\mathrm{dom}\to\mathrm{cod}$.
Then $\tau\circ\Phi$ is again a natural transformation, $\mathrm{dom}\circ \Phi\to\mathrm{cod}\circ\Phi$.
Using again the condition $\phi=\tau\circ\Phi$, we must have
(*) $\quad\mathrm{dom}\circ\Phi=F\$ and $\ \mathrm{cod}\circ\Phi=F'$, $\ \$ because $\phi:F\to F'$.
Now, if $\beta:b\to b'$ is given, $\Phi(\beta)$ is a commutative square in $\mathcal C$, its top and bottom are $\phi_b$ and $\phi_{b'}$, respectively, and (*) says that the left and right sides are $F(\beta)$ and $F'(\beta)$.