Is the algebra map of the ultrafilter monad continuous? Let $\beta$ be the ultrafilter functor from Sets to Sets, which sends a set $X$ to the set of all ultrafilters on the powerset of $\mathcal{P}(X)$ equipped with its Boolean algebra structure.
Then $\beta$ is a monad. If $A$ is a set equipped with a compact Hausdorff topology, $\mathcal{T}(A)$, then $A$ can be made into an algebra with corresponding action $(A, h: \beta(A)\to A)$ given by sending an ultrafilter on $A$ to its point of convergence under the topology.  Likewise any algebra of the monad can be given a compact Hausdorff topology aligning with the above definition.
Furthermore $\beta(A)$ itself always has a natural compact Hausdorff topology taken by letting the basic open sets be of the form $[B] = \{U \in \beta(X) | B \in U \}$ for every subset $B$ of $A$. 
But if $A$ is compact Hausdorff, is the map $h : \beta(A) \to A$ continuous with respect to the two compact Hausdorff topologies involved? I know that for example the monad multiplication should be continuous because the action $\mu : \beta(\beta(A))\to \beta(A)$ is also, by the monad laws, a monad algebra homomorphism, and thus continuous because monad algebra morphisms for $\beta$ are continuous functions. But it seems that if $A$ is not, like $\beta(A)$, totally disconnected, I have trouble proving it.
Suppose an ultrafilter $F$ converges in $U$. Then $U$ is in $F$. But the converse is not necessarily true: if $U$ is in $F$, $F$ could converge to a point on the boundary of $F$.
 A: Yes.  Indeed, let $U\subseteq A$ be an open subset.  Then $h^{-1}(U)$ is the set of ultrafilters on $A$ whose limit is in $U$.  Now if the limit $h(F)$ of an ultrafilter $F$ is in $U$, then there is a neighborhood $V$ of $h(F)$ whose closure is contained in $U$.  Then $V\in F$, and moreover every ultrafilter which contains $V$ has limit in $\overline{V}\subseteq U$.  Thus the basic open set $[V]\subseteq\beta(A)$ is a neighborhood of $F$ which is contained in $h^{-1}(U)$.  Since $F\in h^{-1}(U)$ was arbitrary, $h^{-1}(U)$ is open.  So the inverse image of an open set under $h$ is open, and so $h$ is continuous.
Note also that this is literally the content of the associative property of an algebra over the ultrafilter monad (so, if you already know that $\beta$-algebras are the same as compact Hausdorff spaces you're done, and conversely this is a key step in proving that $\beta$-algebras are the same as compact Hausdorff space).  The associative property for the structure map $h:\beta(A)\to A$ says that the the diagram
$$\require{AMScd}
\begin{CD}
\beta(\beta(A)) @>{\beta(h)}>> \beta(A)\\
@V{\mu_{\beta(A)}}VV @V{h}VV \\
\beta(A) @>{h}>> A
\end{CD}$$
commutes.  The composition $h\beta(h)$ takes an ultrafilter on $\beta(A)$, pushes it forward to $\beta(A)$ along $h$, and then takes its limit in $A$.  The composition $h\mu_{\beta(A)}$ takes an ultrafilter on $\beta(A)$, computes its limit in the topology of $\beta(A)$ (that's what $\mu$ does), and then takes the image under $h$.  So to say that these two maps are equal is to exactly say that the map $h$ preserves convergence of ultrafilters, with respect to the topologies on $\beta(A)$ and $A$.  That is, $h$ is continuous.
In fact, this is similarly true for any monad.  If $T$ is a monad on some category and $h:T(A)\to A$ is a $T$-algebra, then $T(A)$ has a canonical $T$-algebra structure via $\mu_{T(A)}$ and the associativity axiom for $h$ says exactly that $h$ is a $T$-algebra homomorphism from $T(A)$ to $A$.
