The category of functors if the codomain category has zero object is non-empty. Why?

Here, page 6, Daniel Murfet said, the category of functors $$(\mathcal{A}, \mathcal{B})$$ can be empty, although it is nonempty if $$\mathcal{B}$$ has zero object. Why?

[Zero object: An object which is both initial and terminal. ]

• If $\mathcal B$ has a zero object $0$, the constant functor $c_0$, mapping all things to $\mathcal B$'s zero-object is pre-additive. Note that by $(\mathcal A, \mathcal B)$ denotes the category of pre-additive functors. – martini Oct 21 '18 at 14:56
• @martini 1. $\mathcal{B}$ is not preadditve it has just zero object which means it is enriched over the category of monoids. 2. By the same analogy $(\mathcal{A}, \mathcal{B})$ for any two categories should not be empty because you can always have constant functors, assuming $\mathcal{B}$ is not an empty category. – Mobius Knot Oct 21 '18 at 16:12
• I think you might be thinking more is being said than is really being said. The author is just mentioning a common, sufficient condition for this category to be non-empty. Also, just because it seems like this might cause you trouble if you continue reading this paper, that sentence (which is on page 7) does assume that $\mathcal{A}$ and $\mathcal{B}$ are preadditive, and the notation $(\mathcal{A},\mathcal{B})$ has been defined as the category of additive functors between preadditive categories, so talking about $(\mathcal{A},\mathcal{B})$ between arbitrary categories is weird here. – Malice Vidrine Oct 21 '18 at 18:50