Are sheaves (co)algebras for a (co)monad?

Question

I remember reading somewhere that sheafification is a (co)monad on the topos of presheaves. If indeed this is true, then can we characterize the category of sheaves as the Eilenberg-Moore category of (co)algebras for this (co)monad?

Answer 1

Sheafification is a monad simply because the sheafification functor $L$ is left adjoint to the inclusion $i$ of sheaves into presheaves. In fact, this is an idempotent monad: $iLiL\cong iL$ since nothing happens when you sheafify a sheaf. Idempotent monads are equivalent to reflective subcategories, i.e. full subcategories for which the inclusion admits a left adjoint, so this result isn't particularly about sheaves.

Answer 2

Let $Sh(X)$ be a category of sheaves over topological space $X$. For given sheaf $F$ over $X$ and for given point $x\in X$, the stalk of $F$ at $x$, denoted $F_{x}$ is: $F_{x}=\underset{U \ni x }{\varinjlim}F(U)$ the direct limit is indexed over all the open sets containing $x$, with order relation induced by reverse inclusion ( $U<V$, if $U\supset V$). That way we get the functor $J$ to the category $Sets^X$ (here $X$ is viewed as discrete category) or equivalently to the category $Sets/X$ (For given $(S_x)_{x\in X}\in Sets^X $ construct the natural map $\{\underset{x\in X}{\coprod S_x}\to X\}\in Sets/X $, conversely take a preimage of each $x\in X$. That is $Sets^X \cong Sets/X$.)

Now for given map $f:A\to X$ take sections of $f$. The family of all sections form sheaf over $X$. It gives a functor $R$ to $Sh(X)$.

So $G=J\circ R$ is a comonad on the category $Sets^X$ and sheaves are coalgebras over the comonad if it helps you somehow.