Do (special) natural transformations imply a commutative triangle of functors (but NOT vice versa)?

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Given categories $\mathcal{C}, \mathcal{D}$, let functors $F,G: \mathcal{C} \to \mathcal{D}$ be such that

  • for any objects $C_1, C_2 \in Ob(\mathcal{C})$, $F(C_1) = F(C_2) \implies G(C_1) = G(C_2)$,
  • for any morphisms $f_1, f_2 \in Mor(\mathcal{C})$, $F(f_1) = F(f_2) \implies G(f_1) = G(f_2)$.

In this sense the functors $F$ and $G$ are "special". (The latter condition doesn't follow from the first, correct me if I'm wrong.)

(This condition is necessary to write $G$ "as a function of" $F$, as pointed out in this answer to a related question.)

Let $\eta$ be a natural transformation $F \Rightarrow G$, and let $F(\mathcal{C})$ be the image$\dagger$ of $F$, and let $\tilde{F}: \mathcal{C} \to F(\mathcal{C})$ be the functor formed from $F$ by restricting the codomain of $F$ to be $F(\mathcal{C})$.

Claim: The natural transformation induces a functor $H_{\eta}: F(\mathcal{C}) \to \mathcal{D}$ such that $H_{\eta} \circ \tilde{F} = G$, i.e. we have the commutative triangle of functors $$\begin{matrix} \mathcal{C} & \overset{\tilde{F}}{\rightarrow} & F(\mathcal{C}) \\ & \overset{G}{\searrow} & \downarrow H_{\eta} \\ & & \mathcal{D} \end{matrix} $$ However, $H_{\eta}$ by itself is not enough information to recover the natural transformation $\eta: F \Rightarrow G$. In particular, given an arbitrary functor $H: F(\mathcal{C}) \to \mathcal{D}$ such that $H \circ \tilde{F} = G$, it could be induced by any other natural transformation $F \Rightarrow G$, or by no natural transformation at all, because such a functor $H$ by itself doesn't specify any choices of morphisms $F(C) \overset{\phi}{\to} G(C)$ for objects $C \in Ob(\mathcal{C})$, just that each $F(C)$ is "associated with" $G(C)$.

Attempt: Obviously $H_{\eta}$ has to send each $F(C)$ to $G(C)$ for each $C \in Ob(\mathcal{C})$. Likewise $H_{\eta}$ has to send each $F(f)$ to $G(f)$ for every $f \in Mor(\mathcal{C})$. These two assignments should both be well-defined as a result of the "special" condition required of $F$ and $G$ that was given above.

For any morphisms in $Mor(F(\mathcal{C}))$ that are not of the form $F(f)$ for some $f \in Mor(\mathcal{C})$ (but instead are the composition of such morphisms), to guarantee that $H_{\eta}$ can give a well-defined assignment, it seems to be sufficient to show we can ensure $H_{\eta}$ respects composition of morphisms.

So all that remains is to show that $H_{\eta}$ is a functor by showing it respects composition of morphisms and identities.

That $H_{\eta}$ respects identities follows immediately from $F$ and $G$ being functors, so $F(id_C) = id_{F(C)}$ and $G(id_C) = id_{G(C)}$ for any $C \in Ob(\mathcal{C})$, thus given an identity morphism in $F(\mathcal{C})$, which must be of the form $id_{F(C)}$ for some $C \in Ob(\mathcal{C})$, we have that $H_{\eta} ( id_{F(C)}) = H_{\eta}(F(id_C)) = G(id_C) = id_{G(C)} = id_{H_{\eta}(F(C))}$.

Regarding respecting composition - does this follow from the naturality squares defining $\eta$? Or do we not even need to appeal to that at all, making $H_{\eta}$ in fact completely unrelated to the original natural transformation $\eta$?

Comment: Using the identity functor $id_{\mathcal{C}}$ to rewrite the commutative triangle as a commutative square: $$\begin{matrix} \mathcal{C} & \overset{\tilde{F}}{\rightarrow} & F(\mathcal{C}) \\ id_{\mathcal{C}} \downarrow & & \downarrow H_{\eta} \\ \mathcal{C} & \overset{G}{\rightarrow} & \mathcal{D} \end{matrix} $$ writing a natural transformation this way as a commutative square of functors seems related to the ideas discussed on this nLab post. As for the extent to which any of this is really related, I don't know.

$\dagger$ The image $F(\mathcal{C})$ is a category in its own right if we define

  • objects $Ob(F(\mathcal{C}))$ as the collection of objects from $Ob(\mathcal{D}$) of the form $F(C)$ for all $C \in Ob(\mathcal{C})$, and
  • morphisms $Mor(F(C))$ as the "transitive closure" (under composition) of the collection of morphisms from $Mor(\mathcal{D})$ of the form $F(f)$ for all $f \in Mor(\mathcal{C})$.

In particular, although $F(\mathcal{C})$ so defined will be a subcategory of $\mathcal{D}$ albeit usually not a full subcategory.