An endofunctor $F:\mathcal{C} \to \mathcal{C}$ does NOT specify any choices of morphisms $C \to F(C)$, correct?

Question

Given an endofunctor $F:\mathcal{C} \to \mathcal{C}$, although we know that for every object $C \in Ob(C)$ the functor $F$ "sends" $C$ to $F(C)$ in some sense, the functor $F$ by itself does NOT give any choice of morphism in $C \to F(C)$, correct?

In other words, although the endofunctor $F$ declares that each $C$ should be sent to $F(C)$, it still does NOT specify any "implementation" (i.e. via a choice of morphism) for how each $C$ should be sent to $F(C)$. Is that right?

Comments:

1. That seemed to be the point this answer to a related question was making, but then it started discussing absences of maps to the empty set, so I was no longer sure whether that was the point it was making. Although re-reading it, maybe the point of the example is to emphasize why necessarily no choices are being made. Namely the fact that there are even examples where any choice of morphisms "implementing" the assignment is impossible, so (at least for those examples) then necessarily it must be the case that no choices are being made.

2. In this discussion page on nLab, a proposed equivalent definition for functor $F: \mathcal{A} \to \mathcal{B}$ is given that describes $F$ as an endofunctor on $\mathcal{A} \sqcup \mathcal{B}$ and which includes specific choices of morphisms "implementing" each assignment $A \to F(A)$. E.g. $x \overset{\alpha_x}{\to} F(x)$, $y \overset{\alpha_y}{\to} F(y)$. This definition can not be correct, and thus is not equivalent to the standard definition of functor, due to the fact which is the main title of this question, correct?

3. If we did want explicit choices of morphism $C \to F(C)$, we could for example choose some natural transformation $\eta: Id_{\mathcal{C}} \Rightarrow F$ (where $Id_{\mathcal{C}}$ denotes the identity functor for the category $\mathcal{C}$), which would give us explicit choices of morphisms sending $C$ to $F(C)$, i.e. $C \overset{\eta_C}{\to} F(C)$, correct? And given that there might be multiple possible natural transformations $Id_{\mathcal{C}} \Rightarrow F$, this clearly shows that there is no set of unique choices of morphisms $C \to F(C)$ "implementing" the assignment specified by the functor, correct? (Not to mention that we could also choose morphisms "unnaturally" as well.)

Answer 1

Yes, that is correct. Indeed, there need not exist any morphism $C\to F(C)$ at all. For instance, $\mathcal{C}$ could have two objects and only the identity morphisms, and $F$ could swap the two objects (and their identity morphisms).

As you suggest, if you want a functor $F$ to "come with" compatible choices of morphisms $C\to F(C)$, that means you are specifying not just a functor $F$ but also a natural transformation $Id_\mathcal{C}\Rightarrow F$.