I'm having hard time proving this lemma, so hints would be greatly appreciated.
The naturality of collections of isomorphisms is defined as follows:
I was able to prove the right direction. i.e. showing that if the collection is natural, the left square commute iff right square commutes. But I'm having trouble with the reverse direction. It seems hard for me to somehow use an implication, to prove that the isomorphisms are natural.
Hints would be greatly appreciated, thanks!
I’ll only show naturality in $D$, but I hope this helps. Suppose we have $k: d \to d’$. By assumption, we have some isomorphism of sets $D(Fc, d) \cong C(c, Gd)$. Hence consider a morphism $f^{\#}: Fc \to d$. Then this corresponds to some $f^{\flat}: c \to G(d)$, which we may compose with $G(k)$ to obtain $G(k) \circ f^{\flat}: c \to Gd’$.
Now observe that the diagram on the right commutes.
By our assumption the diagram on the left must commute. As a result, we see that $(G(k) \circ f^{\flat})^{\#} = k \circ f^{\#}$, so that $(k \circ f^{\#})^{\flat} = G(k) \circ f^{\flat}$. Hence the diagram commutes.
As Riehl points out, this demonstrates that we have naturality $D$. By a similar arrangement, you can get the other naturality in $C$.