Let's look at the right inverse of a function first:
Let $f:X→Y$, $g:Y→X$ is right-inverse of $f$ (or section of $f$), if only if , $f∘g=id_{Y}.$
It means that If $X$ and $Y$ are finite sets, then $|Y| ≤ |X|$. This because $g$ must be injective and $f$ must be surjective.
Note: If $|Y| > |X|$, then $g$ cannot be the right inverse of $f$.
Analogous to the right inverse of natural transformation:
Let $C$ and $D$ be categories, $S$, $T : C→D$ are functors. If there are two natural transformations $η:S→T$ and $ϵ:T→S$, such that $ϵ∘η=1_{S}$ (e.g. $η$ is the right inverse of $ϵ$), can we come to the following conclusion?
i.e. Pick any pair of objects $\langle c, c'\rangle$ in $C$, the corresponding hom-set in $D$ under $S$ and $T$ must satisfy the following inequality:
|D(Sc, Sc')| ≤ |D(Tc, Tc')|
Note: In category $D$, there may be a lot of morphisms in hom-set $D(Sc, Sc')$. The phrase "corresponding hom-set" means the restriction hom-set under $S$ and $T$.
In particular, if $S$ and $T$ are both injective on objects (i.e. not collapse on objects), then we don't have to use the phrase "corresponding hom-set".
My question is, is this conclusion correct?
The following is my little analysis in the simplest case:
Suppose that
$C$ is a category which has only two objects $c$ and $c'$, only four morphisms $id_{c}:c→c$, $id_{c'}:c'→c'$, $f_{1}:c→c'$, $f_{2}:c→c'$.
$D$ is a category like $Set$.
Functor $S:C→D$ collapses two morphisms $f_{1}$ and $f_{2}$ to one morphism $Sf_{1}$.
Functor $T:C→D$ injects morphism $f_{1}$ to $Sf_{1}$ and $f_{2}$ to $Sf_{2}$.
Now we can say,
It is possible to have two natural transformation $η:S→T$ and $ϵ:T→S$, such that $ϵ∘η=1_{S}$ (i.e. $η$ is right inverse of $ϵ$)
This because two morphisms $Tf_{1}$ and $Tf_{2}$ can be collapsed to one morphism $Sf1$ by commuting square (i.e. $ϵ_{c}'∘Tf_{1}∘η_{c} = Sf_{1} = ϵ_{c}'∘Tf_{2}∘η_{c}$).
But it is impossible to have two natural transformations $η:S→T$ and $ϵ:T→S$, such that $η∘ϵ=1_{T}$ (i.e. $ϵ$ is right inverse of $η$).
This because one morphisms $Sf_{1}$ can not be mapped to two different morphisms $Tf_{1}$ and $Tf_{2}$ at the same time. If it is possible, then $Tf_{1}$ and $Tf_{1}$ must be the same (i.e. $η_{c'}∘Sf_{1}∘ϵ_c = Tf_{1} = Tf_{2}$). But we know $Tf_{1}$ and $Tf_{2}$ are different morphisms.
Can this conclusions be generalized? Or at least this observation makes sense? Or is there any reference about left/right inverse of natural transformation?
Thanks.