Let $X$ be a small site. Let $\aleph$ be an infinite cardinal, such that $|Ob(X)|\leq \aleph$ and $|Mor(X)|\leq \aleph$, where $Mor(X)$ is the set of all morphisms.


We define the size of a presheaf $F$ of sets (or abelian groups, or modules, or ...) as 
$$
|F|=\left|\bigsqcup\_{U\in X} F(U)\right|
$$


In [Flat Covers and Cotorsion Envelopes of Sheaves by Enochs and Oyonarte](https://www.ams.org/journals/proc/2002-130-05/S0002-9939-01-06190-1/S0002-9939-01-06190-1.pdf) the authors mention in passing (on the last page) that if $|F|\leq \aleph$, then $|F^{sh}|\leq \aleph^\aleph$. I was not able to find a reference, but I seem to have an argument for a much stronger bound. I am curious if and where I made a mistake.


For simplicity let us take presheaves of abelian groups for now.


It is well known that the sheafification can be constructed in two steps via the $(-)^\nmid$ funtor defined as
$$
(F^\nmid)(U)=\text{colim}\_{\lbrace U\_i\to U\rbrace}\ker\left(\prod F(U\_i)\to \prod F(U\_i\times\_U U\_j)\right)
$$
Then for any presheaf $F^\nmid$ is separated, and for a separated presheaf $F^\nmid$ is a sheaf. In particular $(F^\nmid)^\nmid$ is always a sheaf.


Now if $|F|\leq \beth$ we can estimate $|F^\nmid|$ as follows. Each kernel can have at most $\aleph\times \beth$ many elements. The colimit is constructed from the direct sum, which can be estimated from the product, and there are at most $2^\aleph\times\aleph=2^\aleph$ many coverings. So we see 
$$
|F^\nmid(U)|\leq 2^\aleph\times \beth
$$
and since there are at most $\aleph$ objects we conclude $|F^\nmid|\leq 2^\aleph\times \beth$. In particular if $\aleph=\beth$ we find $|F^\nmid|\leq 2^\aleph$ and $|F^{sh}|\leq 2^\aleph \times{2^\aleph}=2^\aleph$, which is stronger than what Enochs and Oyonarte claimed.



The size of sheafification
*I'm going to write "$\kappa$" for your "$\aleph$," to more consistently match set-theoretic usage.*


While $2^\kappa$ appears a sharper bound than $\kappa^\kappa$, they are in fact the same (for infinite $\kappa$ at least, and I don't think finite $\kappa$ are important here). $2^\kappa\le\kappa^\kappa$ is clear. For the other direction, we have $\kappa\le 2^\kappa$, so $$\kappa^\kappa\le(2^\kappa)^\kappa=2^{\kappa\times\kappa}=2^\kappa.$$ More generally, whenever $\kappa$ is infinite and $2<\lambda<\kappa$ we have $2^\kappa=\lambda^\kappa=\kappa^\kappa$.


