How to define natural isomorphisms?

Question

Let $\beta(X)$ be the set of all ultrafilters on $X$ and $\beta$ is the ultrafilter . Let $W: CompHaus\to Set^\beta$ be a functor that takes a compact Hausdorff space $X$ to $(X,\xi_X)$ where the map $\xi_X:\beta(X)\to X$ defined where $\xi_X(F)=x$ where $x$ is the limit point of $F$. Let $\hat{W}:Set^\beta\to CompHaus$ be a functor that takes an algebra $(X,\xi_X)$ to a compact Hausdorff space $X$ where we say a set $U$ is open in $X$ if and only if for all $F\in\beta X$ such that $\xi_X(F)\in U$, we have that $U\in F$. How do I define isomorphisms $X\cong \hat{W}(W(X))$ and $(X,\xi_X)\cong W(\hat{W}(X,\xi_X))$ that are natural?

Answer 1

For the first one, note that the underlying set of $X$ is the same as that of $\hat W(W(X))$, so you just have to check that the identity is a homeomorphism, that is, $U$ is open in $X$ if and only if it is in $\hat W(W(X))$

For the second one, in the same way one has that the underlying sets are the same, so one simply has to check that $\xi_X$ is equal to the structure morphism of $W\hat W(X,\xi_X)$

Note that what we get is actually better than natural isomorphisms : we get $W\hat W =id$ as well as $\hat WW = id$, and $\hat W, W$ both commute with the "underlying set" functor, so that the two categories are actually concretely isomorphic (that is, isomorphic as concrete categories), which is a lot stronger than merely equivalent