About one-point compactification

Question

In Folland's "real analysis..." the one-point-compactification is defined for noncompact, locally compact, Hausdorff spaces. (Local compactness is not necessary for the definition, but its one-point-compactification makes Hausdorff). [Hausdorffness also is not necessary, but if we say that space is Hausedorff, than in definition we can use only word "compact" and not "compact closed"]..

There is one exercise in book, page 135, exercise 51: If $X$ and $Y$ are topological spaces and $f$ is a continuous function from $X$ to $Y$, then we say that $f$ is proper if $f^{-1}(K)$ is compact in $X$, for every compact $K \subseteq Y$. Suppose that $X$ and $Y$ are locally compact Hausdorff spaces and $X_1$ and $Y_1$ are their respective one-point-compactifications. If $f$ is continuous function from $X$ to $Y$, than $f$ is proper iff $f$ extends continuously from $X_1$ to $Y_1$ by setting $f(\infty_X)=\infty_Y$.

I've proved that, but nowhere use the fact that $X$ and $Y$ are locally compact (i.e. nowhere use that $X_1$ and $Y_1$ are Hausdorff spaces). Can you tell me whether my proof is correct and local compactness really is not necessary?

1. suppose $F$ is closed in $Y_1$, then:

a. $Y_1\setminus F$ is compact in $Y$ $\Rightarrow$ $f^{-1}[Y_1\setminus F]$ is compact in $X$ $\Rightarrow$ $X_1\setminus f^{-1}[F]$ is compact in $X$ $\Rightarrow$ $f^{-1}[F]$ is closed in $X_1$.

or

b. $F=F_1\cup \{\infty_Y\}$ where $F_1$ is closed in $Y$. It follows that $f^{-1}[F]=f^{-1}[F_1]\cup (f^{-1}[\{\infty_Y\}]=G_1\cup\{\infty_X\}$ where $G_1$ is closed in $X$ $\Rightarrow$ $f^{-1}[F]$ is closed in $X_1$.

2. Suppose $K$ is compact in $Y$ $\Rightarrow$ $K$ is closed in $Y_1$ $\Rightarrow$ $f^{-1}[K]$ is closed in $X_1$ and it does not include $\infty_X$ $\Rightarrow$ $f^{-1}[K]$ is compact in $X$.

Answer 1

As you say, one can define a one-point compactification $X_1 = X \cup \{\infty\}$ for arbitrary spaces $X$ (Alexandroff-compactification). A subset $U_1 \subset X_1$ is defined to be open in $X_1$ if $U_1$ is an open subset of $X$ or $X_1 \setminus U_1$ is a compact closed subset of $X$. In other words, the open neigborhoods of $\infty$ in $X_1$ are the complements of compact closed subsets of $X$.

It is well-known (and very easy to show) that $X_1$ is Hausdorff iff $X$ is locally compact Hausdorff. Note that if $X$ is Hausdorff but not locally compact, then $X_1$ cannot Hausdorff.

The purpose of the above construction is to endow the set $X_1$ with a topology such that

1. The subspace topology of $X$ inherited from $X_1$ is the original topology on $X$.

2. $X$ is open in $X_1$.

3. $X_1$ is compact.

It is easy to see that there exists a unique topology on $X_1$ with these properties. This topology was introduced above. It also explains why we must define open neigborhoods of $\infty$ in $X_1$ to be complements of compact closed subsets of $X$ and not as complements of arbitrary compact subsets of $X$. In fact, if $K \subset X$ is compact and $X_1 \setminus K$ is open in $X_1$, then $K$ is closed in $X_1$ and hence $K = K \cap X$ is closed in $X$.

Note that in general there exist compact subsets $K \subset X$ which are not closed (e.g. in any non-$T_1$ space we find one-point subsets which are not closed). A space in which all compact subsets are closed is called a KC-space. Clearly Hausdorff spaces have this property.

Now we come to your question. To avoid ambiguous notation, let us write $f_1 : X_1 \to Y_1$ for the extension of $f$ via $f_1(\infty_X) =\infty_Y$. We then have without any requirements on $X,Y$

$f_1$ is continuous iff $f$ has property $$(*) \quad f^{-1}(K) \text{ is compact for each compact closed } K \subset Y .$$

$(*)$ is a weaker variant of proper which me may adhoc call cl-proper. It agrees with "proper" if $Y$ is a KC-space, in particular if $Y$ is Hausdorff. Also note that in $(*)$ we can equivalently require that $f^{-1}(K)$ is a compact closed subset of $X$ because $f$ is continuous.

To prove it, consider $V_1 \subset Y_1$ open. If $V_1 \subset Y$, then $f_1^{-1}(V_1) = f^{-1}(V_1)$ is open in $X$ (since $f$ is continuous), and thus open in $X_1$. Therefore $f_1$ is continuous iff $f_1^{-1}(V_1)$ is open in $X_1$ for all $V_1 = Y_1 \setminus K$ with compact closed $K \subset Y$. But $f_1^{-1}(Y_1 \setminus K) = X_1 \setminus f_1^{-1}(K) = X_1 \setminus f^{-1}(K)$, and this set is open in $X_1$ iff $f^{-1}(K)$ is a compact closed subset of $X$.

Therefore you are right, if we work with the Alexandroff-compactification, then it is always true that $f$ is cl-proper iff $f_1$ is continuous. If we insist on your definition of proper, it is true for KC-spaces $Y$, in particular for Hausdorff $Y$.

Let us finally discuss your proof.

Your proof of 2. ($f_1$ continuous implies $f$ proper) is correct, but only works for a KC $Y_1$ since you use that $K$ is closed in $Y_1$. We do not need assumptions on $X_1$ because closed subsets of compact spaces are always compact.

Your proof of 1. ($f$ proper implies $f_1$ continous) doesn't work. You want to show this by proving that if $F_1 \subset Y_1$ is closed, then $f_1^{-1}(F_1) \subset X_1$ is closed. You consider two variants (a.) $Y_1\setminus F_1$ is compact in $Y$ and (b.) $F_1 = F \cup \{\infty_Y\}$. In both cases $\infty_Y \in F_1$, thus you do not consider the case $\infty_Y \notin F_1$, but this is trivial.

In (a.) in general $Y_1 \setminus F_1$ will not be compact, and similarly if $X_1\setminus f^{-1}(F_1)$ is compact in $X$, then in general $f^{-1}(F_1)$ will not be closed in $X_1$.

In (b.) you write $F_1 = F \cup \{\infty_Y\}$ and claim without proof that $f^{-1}(F) \cup \{\infty_X\}$ is closed. But this gap can easily be filled.

Answer 2

Local compactness is necessary for there to be a one-point compactification. For, if $\hat X$ is the one-point compactification of $X$, then $X$ is open in $\hat X$. It follows that, since $\hat X$ is compact Hausdorff, that $X$ is locally compact Hausdorff.

However, it appears we could do without the Hausdorffness assumption, as @BrianMScott points out.