# Extension of a continuous function between topological spaces to their one-point compactification

Let $$X$$ and $$Y$$ be two topological spaces, and let $$X^*$$ and $$Y^*$$ denote their one-point compactification ($$X^* := X \cup \{\infty\},\,\mathcal{T}^* := \{U \subseteq X^*\mid U \cap X \in \mathcal{T} \land (\infty \in U \implies U= (X \setminus K)\cup \{\infty\}$$, $$K$$ closed and compact$$\}$$).

Let $$f:X\rightarrow Y$$ be a continuous function. How can $$f$$ induce a continuous function between $$X^*$$ and $$Y^*$$?

My idea was to define $$f^*(x)=f(x)$$ for $$x\neq \infty$$ and $$f^*(\infty)=\infty$$. Then, let $$U\subset Y^*$$ be an open subset.

• If $$\infty\notin U$$ then $$f^*{}^{-1}(U)=f^{-1}(U)$$ is open since $$f$$ is continuous.

• If $$\infty \in U$$, then $$U=(Y\setminus K)\cup \{\infty\}$$ and $$f^*{}^{-1}((Y\setminus K)\cup \{\infty\})=f^*{}^{-1}(Y\setminus K)\cup f^*{}^{-1}(\{\infty\})=f^{-1}(Y\setminus K)\cup\{\infty\}$$.

But $$f^{-1}(Y\setminus K)=X\setminus f^{-1}(K)$$ and since $$K$$ is closed and $$f$$ is continuous, $$f^{-1}(K)$$ is also closed. Since we know that $$X$$ topological space implies $$X^*$$ compact, and closed subsets of a compact are compact, we conclude $$f^{-1}(K)$$ is compact and therefore $$f^{-1}((Y\setminus K)\cup\{\infty\})$$ is open.

So apparently there seem to be no extra conditions required, however this question (Continuity of the extension of a function between two locally compact Hausdorff spaces to their one point compactifications) suggests otherwise: It requires $$X$$ and $$Y$$ locally compact Hausdorff and $$f$$ proper. Could you suggest where the mistake is?

Thank you very much for your attention.

• For example, what if $X = Y$ is the open unit disk and $f(x)=x/2$? With your definition, $f^*$ is not continuous. Dec 31, 2019 at 17:48

Robert gave you an example. I'll try to explane some "logic".

You said that $$f^{-1}(K)$$ is closed since $$K \subset Y$$ is closed and $$f$$ is continuous. That is true but in the sense of space $$X$$. In the sense of space $$X^*$$ it will be closed if and only if $$f^{-1}(K)$$ is compact in $$X$$ (a closed but not compact subset of $$X$$ is not closed in $$X^*$$). A function that satisfies $$f^{-1}(K)$$ is compact if $$K$$ is compact is called a proper function. It is often referred as "continuity at infinity" (I saw such "definition" in one book about differential topology, I don't remember in which one exactly).

In Robert's example you have a function $$f$$ such that $$f^{-1}(K) = X$$ where $$K = \{z: |z| \le \frac{1}{2}\}$$. $$K$$ is compact in $$Y$$ but $$f^{-1}(K)$$ isn't compact in $$X$$. Therefore $$f$$ is not proper and $$f^*$$ is not continuous.

• Wow, thank you so much! So if I'm not mistaken, if $f$ were proper, then $f^*$ would be continuous, right? Or do we need the hypotheses $X,Y$ locally compact and Hausdorff (as suggested in the other post)? What about if $f$ were not proper, are there any conditions in terms of $X,Y$ such that $f^*$ (or another function induced by $f$) be continuous? I think that $X$ compact would work, but it seems excessive since we are constructing its compactification. Thanks a lot for your help!
– Oski
Dec 31, 2019 at 19:02

The mistake is in the step "Since we know that $$X$$ topological space implies $$X^∗$$ compact, and closed subsets of a compact are compact, we conclude $$f^{−1}(K)$$ is compact..." We know that $$f^{−1}(K)$$ is closed in $$X$$, but for compactness it would need to be closed in $$X^*$$.

Maybe easier to see on an example: take the complex map $$f:\mathbb C\to\mathbb C, f(z)=e^z$$ and then look at the closed disk $$K=D[0,1]=\{z\in\mathbb C:|z|\le 1\}$$ and its complement $$U=\{z\in\mathbb C:|z|>1\}\cup\{\infty\}$$, which is open (as $$D[0,1]$$ is compact). The inverse image of $$U$$ is $$f^{-1}(U)=\{z\in\mathbb C: \Re(z)>0\}\cup\{\infty\}$$ - and this is not open in $$\mathbb C^*$$. Note that, in that case, your set $$f^{-1}(K)=\{z\in\mathbb C:\Re(z)\le 0\}$$ is closed but not compact.

• Wow, thank you so much! So if I'm not mistaken, if $f$ were proper, then $f^*$ would be continuous, right? Or do we need the hypotheses $X,Y$ locally compact and Hausdorff (as suggested in the other post)? What about if $f$ were not proper, are there any conditions in terms of $X,Y$ such that $f^*$ (or another function induced by $f$) be continuous? I think that $X$ compact would work, but it seems excessive since we are constructing its compactification. Thanks a lot for your help!
– Oski
Dec 31, 2019 at 19:02