Show that the functor $H: J \rightarrow \text{Ab}$ sending each $X$ to the free abelian group generated by the set $X$ is not continuous.
I start by assuming that $H$ is continuous and therefore there's a universal cone $v : D \dot \rightarrow F$ for some $F: J \rightarrow \text{Set}$ where $J$ is small. Let $HY = \Bbb Z^{|Y|}$ for any set $Y$.
Since I'm assuming that $H$ is continuous, that means $Hv_j : \Bbb Z^{|D|} \dot \rightarrow \Bbb Z^{|F_j|}$ is a limiting cone. So for any abelian group $A$ and cone $f_j : A \rightarrow \Bbb Z^{|F_j|}$ there should be a unique $h:A \rightarrow \Bbb Z^{|D_j|}$ such that $Hv \circ h = f$.
But I'm having trouble finding the contradiction to this.
Anyone have any ideas?
Note that in sets, the product of a one element set with itself any amount of times is still a one element set. What happens for free abelian groups of rank $1$?