Mac Lane exercise - Elegant comma category exercise proven by S.A Huq I came across this question whilst reading Mac Lane's "Category Theory for the working mathematician" and it struck me as quite a clever idea:
"Given parallel functors $T,S: D \rightarrow C$, show that a natural transformation $\tau :T \rightarrow S$ is the same thing as a functor $\tau :D \rightarrow (T\downarrow S)$ such that $P\tau=Q\tau=id_D$, where $P$ and $Q$ are the projections $P:(T\downarrow S)\rightarrow D$ and $Q:(T\downarrow S)\rightarrow D$."
I really want to prove this, however I am getting nowhere!  I am familiar with all the concepts needed (comma categories, natural transformations etc.) but I just can't seem to find a starting point and I would really appreciate a little insight into how to solve the problem.
Thanks in advance!
 A: Conceptually, the crucial point is this:

Let $U : \mathcal{A} \to \mathcal{C}$ and $V : \mathcal{B} \to \mathcal{C}$ be functors and let $P : (U \downarrow V) \to \mathcal{A}$ and $Q : (U \downarrow V) \to \mathcal{B}$ be the respective projections. Then, there exists a natural transformation $\theta : U P \Rightarrow V Q$ such that, for all functors $F : \mathcal{D} \to \mathcal{A}$ and $G : \mathcal{D} \to \mathcal{B}$ and natural transformations $\alpha : U F \Rightarrow V G$, there exists a unique functor $H : \mathcal{D} \to (U \downarrow V)$ such that $P H = F$, $Q H = G$, and $\theta H = \alpha$.

The exercise you describe is the special case where $\mathcal{A} = \mathcal{B} = \mathcal{D}$ and $F = G = \mathrm{id}_\mathcal{D}$.
A: Hint: Suppose $\tau\colon T \to S$ is a natural transformation, and let $d$ a $D$-object. We want to define its image in $(T \downarrow S)$, define it as $\tau^*d := (d, d, \tau_d)$ (note that $\tau_d \colon Td \to Sd$ is a morphism of $C$). For a morphism $f \colon d \to d'$ define $\tau^*f \colon (d,d,\tau_d) \to (d', d', \tau_{d'})$ as $(f,f)$. Show that $\tau^*$ has the desired properties.
For the other way round, follow our above idea, to define $\tau\colon T \to S$ given a functor $F \colon D \to (T \downarrow S)$, let $\tau_d$ be the morphism $g$  in $Fd = (d,d, g)$.
A: For a given object $d \in D$, the $d$-th component of the natural transformation $\tau$ is an arrow $\tau_d: T(d) \to S(d)$ in $C$. We see that $(d, d, \tau_d)$ is an object in the comma category. This defines the map $\tau^*: D \to (T \downarrow S)$ on objects. (I use $\tau^*$ just to distinguish it from the original natural transformation.)
Naturality of $\tau$ is expressed as follows.
For any morphism $\varphi:d \to d'$ in D, we have a commutative diagram
$$\require{AMScd}
\begin{CD}
T(d) @>{\tau_d}>> S(d)\\
@V{T(\varphi)}VV @VV{T(\varphi)}V \\
T(d') @>{\tau_{d'}}>> S(d')
\end{CD}
$$
It is natural to define $\tau^*(\varphi) = (\varphi, \varphi)$. It is easy to verify that $\tau^*$ is in fact a functor.
It is obvious to see that $P\tau^*(d) = P((d, d, \tau_d)) = d$ and $Q\tau^*(d) = Q((d, d, \tau_d)) = d$ when $P$ and $Q$ are projections of the first component and the second component, respectively.
Also, these are equalities of functors if we define $P$ and $Q$ on morphisms by $P((\alpha, \beta)) = \alpha$ and $Q((\alpha, \beta)) = \beta$. Therefore, $P\tau^* = Q\tau^* = id_D$ as functors.
This proves that $\tau$ induces $\tau^*$.
To show the converse, suppose a functor $\tau^*: D \to (T \downarrow S)$ is given.
The condition $P\tau^* = Q\tau^* = id_D$ implies that
for $d \in D$, $\tau^*(d) = (d, d, \varphi)$ where $\varphi:T(d) \to S(d)$
is a morphism in $C$.
We can then define a natural transformation $\tau$ by
$\tau_d = R\tau^*(d)$, where $R$ is the projection of the third component.
What is left to show is that $\tau$ is natural.
Since $\tau^*$ is a functor,
it maps a morphism $\varphi: d \to d'$ in $D$ to
a pair of morphisms $(\alpha, \beta)$ such that
$$\require{AMScd}
\begin{CD}
T(d) @>{\tau_d}>> S(d)\\
@V{T(\alpha)}VV @VV{S(\beta)}V \\
T(d') @>{\tau_{d'}}>> S(d')
\end{CD}
$$
commutes.
We use the condition $P\tau^* = Q\tau^* = id_D$ on morphisms to derive that $\alpha = \varphi$ and $\beta = \varphi$.
This diagram now coincides with the first diagram above, hence $\tau$ is a natural transformation.
