Seeking more information regarding the "hybriation function." 
Definition 0. Given a pair of finite sets $Y$ and $X$, write $Y_X$ for the set of all collections $\mathcal{K}$ of functions $f : Y \leftarrow X$ that are closed under "hybridization", by which I mean that for all $A,B \in \mathcal{P}(X)$, we have that if $A \cap B = \emptyset$ and $A \cup B = X$, then for all $g,f \in \mathcal{K}$, it holds that the hybrid function $(g \restriction_B) \cup (f \restriction_A)$ is an element of $\mathcal{K}$.

(Please comment if my meaning is unclear.)

Definition 1. Given natural numbers $b$ and $a$, define that $b_a$ equals the natural number $|Y_X|,$ for any and hence all sets $Y$ and $X$ satisfying $|Y|=b$ and $|X|=a$.
This defines a function $$\left(\mathop{\lambda}_{b,a:\mathbb{N}}b_a\right) : \mathbb{N} \leftarrow \mathbb{N} \times \mathbb{N}.$$
Lets call this the "hybriation function."

Now it seems reasonable to think that the hybriation function grows pretty fast in each argument. There's trivially an upper bound, though: $$\mathop{\forall}_{b,a:\mathbb{N}} \left(b_a \leq 2^{b^a}\right).$$
Aside from that, I haven't been able to show much about it.

Questions.
  
  
*
  
*Does the hybriation function have a standard name,
  
*is it known to have any interesting properties,
  
*and can it be usefully expressed in terms of the usual arithmetical operators like $+,\times$, $\sum$ and $\prod$ etc.?
  

 A: Let $\mathcal K$ be a fixed closed collection. For all $x\in X$ define $R(x)\subset Y$ as $\{f(x)|f\in\mathcal K\}$ and set
$$\mathcal L=\{g:X\to Y|\forall x\in X,g(x)\in R(x)\}.$$
Obviously $\mathcal L$ is closed and $\mathcal K\subset\mathcal L.$ We claim that actually $\mathcal K=\mathcal L.$ For if $g\in\mathcal L$ then we have a family of functions $(f_x)_{x\in X}$ all belonging to $\mathcal K$ such that $f_x(x)=g(x)$ for all $x.$ Number the elements of $X=\{x_1,\ldots,x_a\}$ and set $g_1=f_{x_1}$ and recursively $g_{n+1}=(g_n\restriction_{\{x_1,\ldots,x_n\}})\cup (f_{x_{n+1}}\restriction_{\{x_1,\ldots,x_n\}^c}).$ Then all the $g_i$ belong to $\mathcal K$ and $g_a=g.$
Thus we can enumerate the distinct collections $\mathcal K$ that make up $Y_X$ by specifying a choice of $R(x)\subset Y$ for every $x\in X.$ The choices of the $R(x)$ are independent except that if one of them is nonempty, all of them are. That leaves $(2^b-1)^a+1$ possibilities.
Verification for small values of $a$ and $b:$
If $a=0$ then the only function from $X$ to $Y$ is the empty function. The closed collections are the empty collection and the singleton containing the empty function. Our formula is correct even for $a=b=0$ if we adopt the convention $0^0=1.$
If $b=0$ and $a>0$ then there are no functions from $X$ to $Y$ so the only closed collection is the empty collection.
If $a=1$ then there are $b$ functions from $X$ to $Y$ and all $2^b$ collections of them are closed.
If $b=1$ then there is exactly one function from $X$ to $Y$ and there are two closed collections: the empty collection and the singleton containing the one function.
