How does the product construction of the Stone-Cech compactification work?

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Wikipedia's article on the Stone-Cech compactification gives several constructions of it, one which is this:

One attempt to construct the Stone–Čech compactification of $X$ is to take the closure of the image of $X$ in $${\displaystyle \prod C}$$ where the product is over all maps from $X$ to compact Hausdorff spaces $C$. This works intuitively but fails for the technical reason that the collection of all such maps is a proper class rather than a set. There are several ways to modify this idea to make it work; for example, one can restrict the compact Hausdorff spaces $C$ to have underlying set $P(P(X))$ (the power set of the power set of $X$), which is sufficiently large that it has cardinality at least equal to that of every compact Hausdorff set to which $X$ can be mapped with dense image.

Forget the caveat about sets and proper classes, I'm just trying to understand the idea. My question is, what does ${\displaystyle \prod C}$ mean? I don't understand "the product is over all maps from $X$ to compact Hausdorff spaces $C$". What exactly are we taking a product of?

Are we taking a Cartesian product of compact Hausdorff spaces, or are we taking some kind of product of maps, or what?

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Let $S$ be the set* of all continuous maps $f:X\to C$ where $C$ can be any compact Hausdorff space. Then the product in question is the product $$\prod_{f\in S}\operatorname{codomain}(f).$$ That is, the index set of the product is $S$, and the factor corresponding to each map $f:X\to C$ in $S$ is the space $C$. This is just an ordinary product of topological spaces.

Denoting this product by $P$, there is a canonical map $F:X\to P$. Namely, for each $f\in S$, the $f$-coordinate of $F$ is just $f$. That is, $F(x)(f)=f(x)$. The Stone-Cech compactification of $X$ is then (modulo set-theoretic issues) the closure of the image of this map $F$.

*OK, it's actually only a class, but that's not important here.