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Let $X$ be a space and $h: X \to Z$ be an embedding of $X$ in the compact Hausdorff space $Z$. There exists a corresponding compactification $Y$ of $X$ such that $H:Y \to Z$ is an embedding and equals $h$ on X. I don't understand the following:

To prove the compactification unique, suppose $Y_i$ is a compactification of $X$ and that $H_i: Y_i \to Z$ is an embedding that is an extension of $h$, for $i=1,2$. Now $H_i$ maps $X$ onto $h(X)=X_0$. Because $H_i$ is continuous, it must map $Y_i$ into $\text{Cl}X_0$ (I don't see how); because $H_i(Y_i)$ contains $X_0$ and is closed being compact, it contains $\text{Cl}X_0$. Hence $H_i(Y_i) =\text{Cl}X_0$ (again, I don't see how the other set inclusion follows). Then, $H_2^{-1} \circ H_1$ is a homeomorphism of $Y_1$ with $Y_2$.

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  • $\begingroup$ So, what is the definition of a compactification of $X$? $\endgroup$ – Ted Shifrin Aug 27 '15 at 16:36
  • $\begingroup$ @TedShifrin It is a compact Hausdorff space such that $Y$=Cl$X$ $\endgroup$ – Mark Aug 27 '15 at 16:42
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Since $H_i$ is continuous we know that $H_i^{-1}(ClX_0)$ is a closed set which contains X. Since $Y_i$ is the closure of X, we know that $Y_i \subset H_i^{-1}(ClX_0)$. Therefore $H_i(Y_i) \subset ClX_0$.

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  • $\begingroup$ I see. Thanks. And the reverse inclusion? $\endgroup$ – Mark Aug 27 '15 at 16:55
  • $\begingroup$ Similar, since $Y_i$ is compact its image in Z is closed. By definition of the closure of a set we know that $ClX_0 \subset H_i(Y_i)$. $\endgroup$ – Michael Menke Aug 27 '15 at 16:57
  • $\begingroup$ Yes, that is what I was just going to write, following your answer. Thanks so much for your help. $\endgroup$ – Mark Aug 27 '15 at 17:00

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