Ttdfjt

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
$$
|F|=left|bigsqcup_Uin X F(U)right|
$$

In Flat Covers and Cotorsion Envelopes of Sheaves by Enochs and Oyonarte 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
$$
(F^nmid)(U)=textcolim_lbrace U_ito Urbracekerleft(prod F(U_i)to prod F(U_itimes_U U_j)right)
$$
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 $alephtimes beth$ many elements. The colimit is constructed from the direct sum, which can be estimated from the product, and there are at most $2^alephtimesaleph=2^aleph$ many coverings. So we see
$$
|F^nmid(U)|leq 2^alephtimes beth
$$
and since there are at most $aleph$ objects we conclude $|F^nmid|leq 2^alephtimes beth$. In particular if $aleph=beth$ we find $|F^nmid|leq 2^aleph$ and $|F^sh|leq 2^aleph times2^aleph=2^aleph$, which is stronger than what Enochs and Oyonarte claimed.

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). $2^kappalekappa^kappa$ is clear. For the other direction, we have $kappale 2^kappa$, so $$kappa^kappale(2^kappa)^kappa=2^kappatimeskappa=2^kappa.$$ More generally, whenever $kappa$ is infinite and $2<lambda<kappa$ we have $2^kappa=lambda^kappa=kappa^kappa$.