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Let's consider only small categories for simplicity.

Let $C$ and $D$ be categories, $F_0 : C_0 \to D_0$ be a function and $F_1 : C(x,y) \to D(F_0(x),F_0(y))$ be a family of functions indexed by objects $x,y\in C_0$.

tldr; we have the structure of a functor between two categories but not the properties. (We are asking for it to preserve the domain and codomain for simplicity but I digress.)

Question: Is the pair $\{F_0,F_1\}$ a functor if an only if $F_1$ is natural in $x$ and $y$?

By this I mean there exist natural transformations for all $x$: $C(x,-) \Rightarrow D(F_0(x),F_0(-))$, and for all y: $C(-,y) \Rightarrow D(F_0(-),F_0(y))$.

If the answer is negative but there's a positive answer changing the hypothesis a little bit then it's also welcome.

Thanks in advance!

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    $\begingroup$ You mean natural transformations not isomorphisms? $\endgroup$ Commented Nov 10 at 23:30
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    $\begingroup$ No: you only get a semifunctor (restricted to one-object categories, this is a morphism of semigroups rather than monoids). $\endgroup$ Commented Nov 11 at 0:33

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No, take the monoids $(\{1\},\cdot)$ and $(\{0,1\},\cdot)$ and view them as categories $C,D$ with one object $x$. Then $F_0=id:C_0 \to D_0$ is unique and define $F_1(1) := 0$. Let $\eta: C(x,x)=\{1\} \to D(x,x)=\{0,1\}$ with $\eta(1) = 0$.

Then $\eta$ is natural since $(F_1(1))_*\circ\eta = \eta\circ1_*$ (because $F_1(1) \circ \eta(1) = 0 \circ 0 = 0 = \eta(1\circ 1)$) and $(F_1(1))^*\circ \eta = \eta \circ 1^*$ (because $\eta(1)\circ F_1(1) = 0 \circ 0 = 0 = \eta(1\circ 1)$).

But $F_1(1) \neq 1$ i.e. $(F_0,F_1)$ is no functor $C \to D$.

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