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Is the Stone-Čech compactification of the inverse limit of an inverse system $\left\{ X_{i},f_{ij},I\right\} $ of Tychonoff spaces equal to the limit of the inverse system $\left\{ \beta X_{i},\beta f_{ij},I\right\} $, where $\beta f_{ij}$ is the extension of $f_{ij}$ over $\beta X_{i}$ and $\beta X_{j}$?

Similarly, is the Hewitt realcompactification of the limit of an inverse system $\left\{ X_{i},f_{ij},I\right\} $ of Tychonoff spaces equal to the limit of the inverse system $\left\{ \upsilon X_{i},\upsilon f_{ij},I\right\} $, where $\upsilon X_{i}$ is the realcompactification of $X_{i}$?

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    $\begingroup$ Well, it isn’t even true that compactifications of products are products of compactifications. $\endgroup$
    – user131781
    Nov 4, 2019 at 10:53
  • $\begingroup$ How does this give a contradiction? $\endgroup$ Jan 1, 2020 at 19:53

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Let $X_n$ be $\{k\in\mathbb{N}:k\ge n\}$ and let $f_n:X_{n+1}\to X_n$ be the inclusion map. The inverse limit of the system $\{X_n,f_n,\mathbb{N}\}$ is empty; the limit of the system $\{\beta X_n,\beta f_n,\mathbb{N}\}$ is $\beta\mathbb{N}\setminus\mathbb{N}$. Similarly for $\alpha\in\omega_1$ let $Y_\alpha$ be $\omega_1\setminus\alpha$, with $g_\alpha:Y_{\alpha+1}\to Y_\alpha$ the inclusion map. The inverse limit of the $Y_\alpha$s is empty; as $\upsilon Y_\alpha=[\alpha,\omega_1]$ for all $\alpha$, the inverse limit of the $\upsilon Y_\alpha$s consists of one point, $\omega_1$.

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