Alternative functor construction from universal morphisms

Question

Let $G : \mathcal{D} \rightarrow \mathcal{C}$ be a functor. Suppose that for each object $X \in \mathcal{C}$, there exists a universal morphism $(F_X, \eta_X)$ from $X$ to $G$. The theory of adjoint functors tells us that we can construct a unique functor $F : \mathcal{C} \rightarrow \mathcal{D}$ such that $F(X) = F_X$ and $\eta = (\eta_X)_X$ is a natural transformation. Indeed, given $f : X \rightarrow Y$, we can use universality of $(F_X, \eta_X)$ and define $F(f)$ to be the unique $h : F_X \rightarrow F_Y$ such that $G(h) \circ \eta_X = \eta_Y \circ f$. This $F$ will be a left adjoint of $G$.

However, it seems plausible to construct another functor $F'$ such that $F'(X)=F_X$, but which is defined differently on arrows, in a way so that $F'$ is no longer a left adjoint of $G$. To get around the above uniqueness condition, it would have to be done so that $\eta$ is no longer a natural transformation. Is this possible? Are there any simple examples (e.g. where $G : Grp \rightarrow Set$ is the forgetful functor and $F : Set \rightarrow Grp$ is the free functor)?

EDIT: I realized my question is equivalent to the following. Let $F,F' : \mathcal{C} \rightarrow \mathcal{D}$ and $G : \mathcal{D} \rightarrow \mathcal{C}$ be functors. Suppose that $F$ and $F'$ agree on objects (i.e. $F(X)=F'(X)$ for each object $X \in \mathcal{C}$). If $F$ is a left adjoint of $G$, does it follow that $F'$ is also a left adjoint of $G$?

Answer 1

Let $F,F' : \mathcal{C} \rightarrow \mathcal{D}$ and $G : \mathcal{D} \rightarrow \mathcal{C}$ be functors. Suppose that $F$ and $F'$ agree on objects (i.e. $F(X)=F'(X)$ for each object $X \in \mathcal{C}$). If $F$ is a left adjoint of $G$, does it follow that $F'$ is also a left adjoint of $G$?

In the special case $F = G = \mathrm{id}_{\mathcal{C}}$, this becomes: If $F ' : \mathcal{C} \to \mathcal{C}$ is a functor which is the identity on objects, is $F' \cong \mathrm{id}_{\mathcal{C}}$? This would imply in particular that $F'$ is fully faithful. But this already fails for one-object categories, i.e. monoids. Any non-bijective monoid homomorphism $ M \to M$ is a counterexample.

Answer 2

Yes, this is possible.

Let’s start with an extremely simple example: the identity functor $Set \to Set$, whose adjoint is itself. Fix a set $S$ and a nontrivial permutation $\pi : S \to S$. We can define a functor $F : Set \to Set$ by $F(X) = X$ for all objects $X$, and for all morphism $f : X \to Y$,

$$F(f) = \begin{cases} f & X \neq S \neq Y \\ f \circ \pi & X = S \neq Y \\ \pi^{-1} \circ f & X \neq S = Y \\ \pi^{-1} \circ f \circ \pi & X = S = Y \end{cases}$$

We have a natural isomorphism $\theta : F \to 1_{Set}$ given by

$$\theta_X = \begin{cases} 1_X & X \neq S \\ \pi & X = S \end{cases}$$

Thus, we see that $F$ is the same as $1_{Set}$ on objects, and the two functors are isomorphic, yet they are not the same on arrows.

In fact, we could start with any category $C$ and any choice $(\theta_X : X \to X)_{X \in C}$ of isomorphisms that is nontrivial and get a similar result by defining $F(f) = \theta_Y^{-1} \circ f \circ \theta_X$. So there are plenty of examples.