Show the existence of a particular map from the compactification of a space to its one-point compactification.

Question

I'm trying to do the following exercise.

I've been looking around but have not seen the definition of the map i. Is it just an open embedding or is it something else? What are its properties?

Also, I plan to proceed with the proof by assuming that f exists and then finding its definition. If $k \in K$ is such that $k = e(x)$ for some $x \in X$, and if i is indeed an open embedding, then can I just simply state $f(k)$ to be such that $i(x) = f \circ e(x) = f(k)$ holds? What about the other case where $k \notin e(X)$?

Answer 1

I assume (you should really check exercise sheet 4 for this) that $i(x)=x$ and $X^+ = X \cup \{\infty\}$ for some point $\infty \notin X$, topologised in the usual way (all open sets of $X$ are also open in $X^+$ and we add all subsets of $X^+$ that contain $\infty$ and whose complement is a compact subset of $X$) and this implies in particular that $i(x)=x$ from $X$ to $X^+$ is an open embedding.

It's clear that you have no choice in defining $f$ on $e[X]$: $f(y)=x$ when $y=e(x) \in e[X]$ and then define $f(y)=\infty$ otherwise (i.e. $y \in K\setminus e[X]$). Check that this is continuous, which is easy with the topology on $X^+$ as described above. Also, $e[X]$ is dense in $K$ so once we've defined $f$ on that, it is uniquely determined (as $X^+$ is Hausdorff), by a standard theorem.

Answer 2

Hints:

$1).\ e(X)$ is open in $K$.

$2).\ $ Let $\infty$ be the point added to $X$ to obtain $X^+$ and define $f:K\to X^+$ by

$ f(k)=\begin{cases} e^{-1}(k) & k\in e(X)\\ \infty & k\in K\setminus e(X) \\ \end{cases}$

$3).\ $ show $f$ is continuous, using $1).$

edit:

$1).$ is the hardest step, and follows from the fact that if $A\subseteq X$ is locally compact and $X$ is Hausdorff, then $A$ is open in $\overline A$.

Indeed, let $a\in A$ and $N_a$ be a neighborhood of $a$ (in $A$), such that cl$_A N_a:=K$ is compact. Then, there is an open set $V$ in $X$ such that $A\cap V=N_a.$

The foregoing implies that $\overline{A\cap V}\cap A=\overline N_a\cap A=$cl$_A N_a=K$ so $\overline{A\cap V}\cap A$ is compact and therefore closed in $X$.

But $A\cap V\subseteq \overline{A\cap V}\cap A$ and we just showed that the RHS of this is closed, so $\overline {A\cap V}\subseteq \overline{A\cap V}\cap A$ and so $\overline {A\cap V}\subseteq\overline N_a\cap A\subseteq A$.

And since $V\cap\overline A\subseteq \overline {V\cap A}$, (because $V$ is open in $X$), it follows that $V\cap\overline A$ is contained in $A$, contains $a$, and is open in $\overline A$. This finishes the proof.