Limit along the category of all algebraic curves over a field

Question

Let $k$ be algebraically closed field of charactersistic zero and $\mathcal C$ be the category of irreducible smooth projective curves over $k$ and non-constant maps between them. I have a functor $F\colon \mathcal C\to\ Aff$ ($Aff$ is the category of all affine spaces over $\mathbb Q$; they can be infinite dimensional). I know that:

1. For any $X\in\mathcal C$ the affine space $F(X)$ is not-empty,
2. If curve $X$ is elliptic, hyperelliptic or rational then $|F(X)|=1$.

Is it possible to deduce that $\lim\limits_{\mathcal C}F$ is not empty? If it is not true, then what additional conditions I need to assume?

Let us assume that I know that for any $f\colon X\to Y$ the corresponding map $F(f)$ is surjective. Does it change anything?

My motivation comes from the following example of $F$. In the following all abelian group are supposed to be tensored by $\mathbb Q$. The map $\Lambda^3 k(X)\to B_2(k)$ (where $B_2(k)$ is so-called pre Bloch group of $k$) is called strong reciprocity law if the following diagramm is commutative: $$\require{AMScd} \begin{CD} B_2(k(X))\otimes k(X)^\times @>{\delta_3}>> \Lambda^3k(X)^\times\\ @VV{\sum\limits_{x\in X(k)}\partial_x}V@VV{\sum\limits_{x\in X(k)}\partial_x}V \\ B_2(k) @>{\delta_2}>> \Lambda^2k^\times \end{CD}$$

Here the map $\partial_x$ is so-called time symbol, $\delta_3$ is given by the formula $\delta_3(\{f\}_2\otimes g)=f\wedge (1-f)\wedge g, \delta_2(\{f\}_2)=f\wedge (1-f)$. Details can be found in part 6 of

Goncharov, A. B., Polylogarithms, regulators, and Arakelov motivic complexes, J. Am. Math. Soc. 18, No. 1, 1-60 (2005). ZBL1104.11036.

Define $F(X)$ to be the set of all strong reciprocity laws on $X$. If $X\to Y$ is non-constant morphism and we have an element $h\in F(X)$, we can define $F(f)(h)$ as the composition of the maps $f^*\colon \Lambda^3k(Y)^\times\to \Lambda^3k(X)^\times$, $h\colon \Lambda^3 k(X)^\times\to B_2(k)$ devided by degree of $f$. One can prove that $F(f)(h)$ is a strong reciprocity law on $Y$.

According to result of D. Rudenko the set $F(X)$ is non empty for any $X$. The problem is whether there exist a strong reciprocity law $h_X$ on any curve $X$ such that for any non-constant $f\colon X\to Y$ it holds that $F(f)(h_X)=h_Y$? I know how to prove it, but the proof is complicated. So I want to know whether it can be proven by a simple category-theoretical arguments or not.

Answer 1

Here is a counterexample. Put $F(X)=\mathbb{Q}$ for all $X$, and for every morphism $f:X\to Y$ put $F(f)=$ multiplication by $\deg(f)$.

Now assume $V$ is a $\mathbb{Q}$-vector space and $\phi:\underline{V}\to F$ is a morphism, where $\underline{V}$ is the constant functor. Thus for each $X$ we have a map $\phi(X): V\to \mathbb{Q}$, satisfying $\phi(Y)=\deg(f)\phi(X)$ for each $f:X\to Y$. It follows that $\phi(\mathbb{P}^1)=0$ (consider any $f:\mathbb{P}^1\to\mathbb{P}^1$ of degree $>1$). Since any $X$ admits a morphism $g:X\to\mathbb{P}^1$ we conclude that $0=\deg(g)\phi(X)$, i.e. $\phi(X)=0$.