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Is it true that the Stone-Čech compactification preserves disjoint union (even infinite disjoint unions)? That is, is it true that $\beta(\bigsqcup_\alpha X_\alpha) = \bigsqcup_\alpha \beta X_\alpha$?

This seems to me true since the Stone-Čech compactification functor is left adjoint to the inclusion functor (from the category of topological spaces into the category of compact Hausdorff topological spaces), and we know that left adjoint functors preserve all colimits (as discussed here).

Yet, if $X$ is a discrete then we can write $X=\bigsqcup_{x \in X} \{x\}$ but $\beta X \neq \bigsqcup_{x \in X} \{x\}$ (where $\beta(\{x\}) = \{x\}$)

Serge the Toaster
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Yes, it's true that the Stone-Čech functor $\beta : \mathsf{Top} \to \mathsf{CompHaus}$ is left adjoint to the inclusion $\iota : \mathsf{CompHaus} \hookrightarrow \mathsf{Top}$. Thus it preserves colimits, and in particular it's true that

$$ \beta \left ( \coprod_{x \in \mathbb{N}} {}^{\mathsf{Top}} \{ x \} \right ) = \coprod_{x \in \mathbb{N}} {}^{\mathsf{CompHaus}} \beta \{ x \} = \coprod_{x \in \mathbb{N}} {}^{\mathsf{CompHaus}} \{ x \} $$

However, as I've made explicit in the above notation, the second coproduct is computed in $\mathsf{CompHaus}$, not in $\mathsf{Top}$!

Since $\mathbb{N}$ is, among other things, not compact hausdorff, the coproduct $\coprod_{x \in \mathbb{N}}^{\mathsf{CompHaus}} \{ x \}$ cannot possibly be $\mathbb{N}$. So it's possible that this issue goes away, and indeed it does.

What your computation has secretly shown is that, in the category of compact hausdorff spaces, the coproduct of countably many singletons is $\beta \mathbb{N}$!

In general, this is one way to compute colimits in a reflective subcategory. We compute colimits in the big category (for us, $\mathsf{Top}$) then apply the reflector (for us $\beta$). You can see someone asked a similar question here.

I hope this helps ^_^

HallaSurvivor
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  • That is a great answer, thanks! So, if I understand correctly, I would have to replace the index on which I take the disjoint union in order to compute the disjoint union in the category $CompHaus$ ? – Serge the Toaster Jan 21 '23 at 07:55
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    @SergetheToaster No, the index doesn't change. $\coprod_{i\in I}^{\mathsf{CompHaus}} X_i$ is the given of a compact T2 topological space $X$ and continuous maps $\iota_i:X_i\to X$ such that for all compact T2 topological space $Y$ and all continuous maps $t_i:X_i\to Y$ there is exactly one continuous map $r:X\to Y$ such that $\forall i, t_i=r\circ \iota_i$. – Sassatelli Giulio Jan 21 '23 at 10:16
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    @SergetheToaster Compare with $\coprod_{i\in I}^{\mathsf{Top}} X_i$ being the given of a topological space $X$ and continuous maps $\iota_i:X_i\to X$ such that for all topological space $Y$ and all continuous maps $t_i:X_i\to Y$ there is exactly one continuous map $r:X\to Y$ such that $\forall i, t_i=r\circ \iota_i$. – Sassatelli Giulio Jan 21 '23 at 10:19
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    @SergetheToaster In the case of singletons, the discrete topological space is not a coproduct in $\mathsf{CompHaus}$ because it isn't compact T2, despite satisfying the condition on morphisms. Their coproduct in $\mathsf{CompHaus}$ is not a coproduct in $\mathsf{Top}$ because the condition on morphisms for $\mathsf{CompHaus}$ is less restrictive than it is in $\mathsf{Top}$. – Sassatelli Giulio Jan 21 '23 at 10:25
  • @SassatelliGiulio I see. Many thanks for the explanations! – Serge the Toaster Jan 21 '23 at 17:43