Functoriality of the one-point compactification

Question

My question concerns the passage below: Why does continuity at infinity of $\phi$ imply that the extension $\phi^{+}$ to the one-point compactifications necessarily maps $\infty_{X}$ to $\infty_{Y}$? (NB A continuous map between locally compact Hausdorff spaces is continuous at infinity if it extends to a continuous map between their one-point compactifications.) For instance, the constant map $c_{0}\colon \mathbb{R} \to \mathbb{R}$ taking the value $0$ at all real numbers is continuous at infinity, but I believe its extension $c_{0}^{+}$ must map $\infty_{\mathbb{R}}$ also to $0$.

E. Park. Complex Topological K-Theory. Cambridge Studies in Advanced Mathematics. Cambridge University Press, 2008.

Addendum (Definition of continuous at infinity.)

Answer 1

Based on Definition 2.6.2 the Proposition is false if the morphisms of the category of locally compact Hausdorff spaces are understood as the maps which are continuous at infinity.

There are two ways to correct it:

1. The assignment $X \mapsto X^+$ defines a covariant functor from the category of locally compact Hausdorff spaces and maps which are continuous at infinity to the category of compact Hausdorff spaces.

Here no basepoints are involved, one does not require that the extension $\phi^+$ of $\phi : X \to Y$ satisfies $\phi^+(\infty_X) = \infty_Y$.

2. The assignment $X \mapsto (X^+,\infty_X)$ defines a covariant functor from the category of locally compact Hausdorff spaces and proper maps to the category of pointed compact Hausdorff spaces.

See Is one point compactification functorial?

The problem with the orginal Proposition is that there are maps which are continuous at infinity but not proper (for example constant maps), though of course proper maps are continuous at infinity.

Remark.

The author's intention is to extend a map $\phi : X \to Y$ to a continuous $\phi^+ : X^+ \to Y^+$ such that $\phi^+(\infty_X) = \infty_Y$. This works if and only if $\phi$ is proper.

Note that if $X$ is compact, then all maps $\phi : X \to Y$ are proper and thus can be extended via $\phi^+(\infty_X) = \infty_Y$. However, in this case there are more continuous extensions of $\phi$. In fact, $\infty_X$ (which ís an isolated point of $X^+$) can be mapped to any $y \in Y$. But these additional extensions are irrelevant for the purpose of finding a functor.

In 1. the only case in which $\phi^+(\infty_X) = \infty_Y$ cannot be achieved is when $\phi$ is continuous at infinity, but not proper. Here we have $\phi^+(\infty_X) \in Y$. Recall that for compact $X$ there are no non-proper $\phi$.