Show that $F$ cannot be a extended to morphisms of category $\mathcal{Set}$

Question

Let $F(X) = X \cap \mathbb{N}$. I have to show that $F$ cannot be extended to morphisms of category $\mathcal{Set}$. I wanted to assume it can and then find a contradiction with on of the properties of functors (i.e. $F(g \circ f) = F(f) \circ F(g)$ and $F(id_X) = id_{F(X)}$) but it seems like I need something more to do it. How can I show this knowing only how the functor should transform objects?

Answer 1

You hardly need any properties of functors at all to show that such a functor $F \colon {\mathcal Set} \to {\mathcal Set}$ cannot exist.

Consider a set $X$ with non-empty intersection with ${\mathbb N}$, a non-empty set $Y$ with empty intersection with ${\mathbb N}$, and a function $f \colon X \to Y$. Now where would $F$ send $f$? The domain of $F(f)$ is $F(X) = X \cap {\mathbb N}$, which is non-empty, and its codomain is $F(Y) = Y \cap {\mathbb N} = \emptyset$.

Answer 2

Your question is not at all about functors. Your $F$ is not required to satisfy your two equations but only to send each map $f:X\to Y$ to some map $F(f):X\cap\Bbb N\to Y\cap\Bbb N.$ And even that few it is impossible if, e.g., $f$ is the constant function from $X=\Bbb N$ to $Y=\{-1\}.$