Product-preserving functors Let $\bf{S}^0$ be the dual category of sets, let $\mathcal{U}_{\infty}$ be a category and $A_{\infty}:\bf{S}^0\longrightarrow\mathcal{U}_{\infty}$ be a functor which is bijective on the object classes. We may identify the objects of $\mathcal{U}_{\infty}$ with the objects in $\bf{S}^0$. For two sets $X$ and $Y$, we define
$$Mor_{\mathcal{U}_{\infty}}(X,Y):=Mor_{\mathcal{U}_{\infty}}(X,1)^Y$$
Then $A_{\infty}$ will become a product-preserving functor.
Questions:
1) what does $Mor_{\mathcal{U}_{\infty}}(X,Y):=Mor_{\mathcal{U}_{\infty}}(X,1)^Y$ mean? I suppose it is a product of morphisms sets, but I would like a more precise description.
2) why do $Mor_{\mathcal{U}_{\infty}}(X,Y):=Mor_{\mathcal{U}_{\infty}}(X,1)^Y$ imply that $A_{\infty}$ is a product-preserving functor?
 A: Firstly, I assume that objects of $U_\infty$ are identifed with objects of $\bf S^0$, i.e. with sets, via the functor $A_\infty$.
I feel some confusion here. The notation $A:=B$ usually means to define the symbol $A$ as $B$. 
Now the category $U_\infty$ seems to be given, though you want to override the homsets.
Its resolution is that either we simply forget the colon and read it as a given equation (or rather, as a natrual isomorphism), or that $U_\infty$ is only partially given beforehand (i.e. only the homsets $\hom_{U_\infty}(X,1)$ are given). 
To answer your questions:


*

*Yes, this is the usual Cartesian power of sets: for sets $A$ and $B$, $\ A^B$ is defined as the set of all functions $B\to A$. (In other words, $A^B=\hom_{\bf Set}(B,A)=\hom_{\bf S^0}(A,B)$). 
So that, $\hom_{U_\infty}(X,1)^Y$ is the set of all functions $Y\to\hom_{U_\infty}(X,1)$. 
Note the analogy that $\hom_{\bf Set}(1,X)\cong X$. In view of this, we might regard morphisms $X\to 1$ in $U_\infty$ as some 'generalized elements' of set $X$.

*Let $Y_\alpha$ and $X$ be sets. Then a cone $(X\overset{f_\alpha}\to Y_\alpha)_{\alpha\in I}\,$ in $U_\infty$ corresponds to the collection of functions $Y_\alpha\to\hom_{U_\infty}(X,1)$, that correspond to a single function from the disjoint union $\coprod_\alpha Y_\alpha\to\hom_{u_\infty}(X,1)$, which corresponds to an element of $\hom_{U_\infty}(X,\,\coprod_\alpha Y_\alpha)$. Note also that $\coprod_\alpha Y_\alpha$ is just the product in $\bf S^0$.

