1
$\begingroup$

A topological space is said to be locally compact if each point $x\in X$ has at least one neighbourhood which is compact. If $f$ is continuous open mapping of a locally compact space $(X,\tau)$ onto a topological space $(Y,\tau_1)$ then $(Y,\tau_1)$ is locally compact.

If $y\in Y$ then there exists a neighbourhood $V$ so that $y\in V$. Suppose there exists at least an $x$ such that $f^{-1}(y)=x$ then $f^{-1}(V)$ contains $U$ that is a compact neighbourhood of $x$. Then $y\in f(U)\subset V$.

In the previous question it was asked:

Prove continuous image of a locally compact space is not necessarily locally compact.

Questions:

1) What is the difference in proof from the present question and the previous one? Why does it change from "not necessarily compact" to "locally compact"?

2) Is my proof right?

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ @Jakobian : Some people define a neighborhood of $x$ to be a set whose interior contains $x$, so their neighborhoods aren't necessarily open sets. I don't care for that myself, I like neighborhoods to be open sets. $\endgroup$ – MPW Dec 27 '18 at 14:52
  • $\begingroup$ @Jakobian No, the definition I wrote is the one the book introduce me to. $\endgroup$ – Pedro Gomes Dec 27 '18 at 14:52
  • $\begingroup$ Note that in the first question, $f$ is also assumed to be open, not just continuous. That's not the case in the second question. $\endgroup$ – MPW Dec 27 '18 at 14:53
  • $\begingroup$ @MPW I know that. What I do not understand is why the fact the function being open changes to the fact that $(Y,\tau_1)$ must be locally continuous. $\endgroup$ – Pedro Gomes Dec 27 '18 at 14:55
  • 1
    $\begingroup$ I guess you mean "locally compact", not "locally continuous"? Anyway, the big thing that openness of $f$ gives you is that if $U$ is a neighborhood of $x$, then $f(U)$ is a neighborhood of $f(x)$. If $f$ is not open, this isn't necessarily true. $\endgroup$ – MPW Dec 27 '18 at 14:59
2
$\begingroup$

As to 1, the difference is that $f$ is now also assumed to be open, when previously we only had continuity. Note that this immediately "kills" the counterexample to that first question, that uses the identity from a discrete space. This map is only open when the image space also has the discrete topology.

Your proof is sloppy and unclear, and does not explicitly use the extra assumption of openness, which is a bad sign for a proof.

Better: let $f: X \to Y$ be continuous open and surjective and $X$ locally compact. Then $Y$ is locally compact: let $y \in Y$ be arbitrary. By ontoness we find $x \in X$ with $f(x)=y$. By local compactness of $X$ there is a compact set $C \subseteq X$ such that $C$ is a neighbourhood of $x$, i.e. there is an open subset $O$ of $X$ with $x \in O \subseteq C$.

Now $$y = f(x) \in f[O] \subseteq f[C]$$

which shows that $f[C]$, which is compact (as $f$ is continuous!) is a neighbourhood of $y$ (as $f[O]$ is open (by openness of $f$!) inside $f[C]$ containing $y$). So $Y$ is locally compact at $y$, and as $y$ was arbitary, everywhere.

Note that we now used all the assumptions on $f$.

$\endgroup$
3
$\begingroup$

1) The second question is asking you for an example of topological spaces $X$ and $Y$ and of a function $f\colon X\longrightarrow Y$ such that:

  • $f$ is continuous;
  • $f$ is surjective;
  • $X$ is locally compact;
  • $Y$ is not locally compact.

2) You did not justify the assertion that $f(U)$ is a neighborhood of $y$. And $f$ doesn't need to have an inverse; therefore, the equality $f^{-1}(y)=x$ makes no sense.

$\endgroup$
  • $\begingroup$ How should I prove the first? How should I write a proof of the first question so that it does not apply to the second one? $\endgroup$ – Pedro Gomes Dec 27 '18 at 14:58
  • $\begingroup$ I write instead $f(x)=y$? How should I justify the assertion $f(U)$ is a neighbourhood of $y$? $\endgroup$ – Pedro Gomes Dec 27 '18 at 15:14
  • 1
    $\begingroup$ You write: take $x$ such that $f(x)=y$. And $f(U)$ is a neighborhood of $y$ because $U$ contains an open set $O$ such that $x\in O$ and $O\subset U$. Therefore, since $f$ is open (you did not use this hypothesis) $f(O)$ is open and $y\in f(O)\subset f(U)$. This proves that $f(U)$ is a neighborhood of $y$. $\endgroup$ – José Carlos Santos Dec 27 '18 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.