Universal property of one-point-compactification

compactificationcompactnessgeneral-topology

In this question, we define locally compact to be Hausdorff and every point has a compact neighbourhood. Similarly, we define compact to be Hausdorff and every open cover has a finite subcover.

Then we have the following universal property:

Let $X$ be a locally compact space, and let $i \colon X \hookrightarrow Y$ be a one-point-compactification, i.e. $i$ is a continuous embedding and $Y = X \cup \{\infty\}$ is compact. Given any continuous map $f \colon X \to Z$ into a compact space $Z$ and a special point $z_0 \in Z$ with the property that $f^{-1}(K) \subseteq X$ is compact for every compact $K \subseteq Z$ with $z_0 \notin K$, there exists a unique continuous lift $\bar{f} \colon Y \to Z$ of $f$.

I know that extending $f$ through $\bar{f}(\infty) = z_0$ is such a lift, I fail to show uniqueness, i.e. what happens if $\bar{f}(\infty) \ne z_0$.

Some facts I know:

  • Finite subsets of Hausdorff spaces are closed.
  • Closed subsets of compact spaces are compact.
  • Compact subsets of Hausdorff spaces are closed.
  • Continuous maps between Hausdorff spaces map compact sets to compact sets.

Best Answer

As stated, this is not correct. It also makes stronger assumptions than it needs to.

Let $X$ be a Hausdorff space but non-compact space. The Alexandroff topology on $Y=X\sqcup\{\omega\}$, where $\omega\notin X$, makes $Y$ a compact but not necessarily Hausdorff space with a continuous embedding $\iota:X\hookrightarrow Y$.

If $Z$ is a Hausdorff but not necessarily compact space and $f:X\to Z$ a continuous function, $z_0\in Z$ a point with $f^{-1}(K)$ being a compact subspace of $X$ for each closed subset of $Z$ not containing $z_0$, then there is a unique continuous $F:Y\to Z$ with $F\circ\iota=f$, and we have $F(\omega)=z_0$.

Your statement is not correct because it allows for $X$ to be compact. If $X$ is compact, then $\omega$ is an isolated point of $Y$ and $F(\omega)$ can be literally anything; uniqueness is lost.

I also allow $Z$ to be non-compact. If $Z$ is compact then the closed sets are precisely the compact sets, so, my statement boils down to your statement. Moreover, since your $Y$ is Hausdorff it has to be homeomorphic to my $Y$, so there are no issues there.

The proof is simple. Since $Z$ is Hausdorff and $\iota(X)$ is dense in $Y$, if $F,F':Y\to Z$ are both extensions of $f$ then their equaliser is closed and contains $\iota(X)$ so it must be the whole space; $F(\omega)=F'(\omega)$ is forced. It reamains to show that $\omega\mapsto z_0$ makes $F$ continuous at the point $\omega$.

That is true if and only if for every open $V\ni z_0$ in $Z$, $F^{-1}(V)=\{\omega\}\cup f^{-1}(V)$ is open in $Y$. That's true if and only if $f^{-1}(V)$ has compact complement in $X$, which is true if and only if $f^{-1}(K)$ is compact where $K:=Z\setminus V$ is a closed set not containing $z_0$. So, it follows more or less by hypothesis.

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