5
$\begingroup$

Those questions about connectivity drive me crazy - I'm having so much difficulty proving them.

Say $X$ is a locally connected space, and $f: X \to Y$ is onto where $Y$ has the quotient topology. Prove that $Y$ is locally connected.

I took $y \in Y$ and went hunting for an open connected neighbourhood $V \subseteq Y$ that contains it. I defined $D := f^{-1}[\{y\}]$. Each $x \in D$ has an open connected neighbourhood that contains it $U_x$.

I looked at $\bigcup_{x \in D}f[U_x]$. It is open in $Y$ since each $f[U_x]$ is connected and all of them contain $y$. So it is a connected set that contains $y$, but it isn't necessarily open - or is it?

I'm not sure how to continue from here...

$\endgroup$
  • $\begingroup$ Actually that set is open. It's because of a particular property of quotient maps (such as $f$). Do you know this property? $\endgroup$ – john Jul 21 '15 at 17:33
  • $\begingroup$ I know that a set $V \subseteq Y$ is open $\iff$ $f^{-1}[V] \subseteq X$ is open. But why should I believe that $f^{-1}[f[\bigcup_{x\in d}U_x]]$ is open? $\endgroup$ – amirbd89 Jul 21 '15 at 17:36
  • 1
    $\begingroup$ Apologies - you are right, that set need not be open. Consider the map $q: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ obtained by taking the quotient by the open unit disk. The preimage of the origin is then the entire open unit disk, and we can choose connected open neighborhoods $U_x$ of every point $x$ in the open unit disk so that $U_x$ itself is still in the open unit disk. In this case, your set (for $y$ the origin in $\mathbb{R}^2$) would just be the point $y$ itself. $\endgroup$ – john Jul 21 '15 at 18:50
  • 1
    $\begingroup$ What you can do instead is, for your chosen $y \in Y$, choose an open neighborhood $V$ of $y$. If it's connected, then great. If not, let $V_y$ be the connected component containing $y$. You want to show that $V_y$ is open and thus is an open connected neighborhood. This is when you want to look at the preimage $f^{-1}(V_y)\subset X$ and use the local-connectedness of $X$. $\endgroup$ – john Jul 21 '15 at 18:53
7
$\begingroup$

Lemma: Let $X$ be a locally connected. If $U$ is an open set in $X$ then all connected components of $U$ are open sets in $X$.

Proof of the lemma: Let $U$ be an open set in $X$ and $W$ one component of $U$. For any $x\in W$, there is an open connected neighbourhood $W_x$ in $X$, such that $W_x\subseteq U$. But since $W_x$ is connected, we have $W_x\subseteq W$. So we have proved that, for any $x\in W$, there is an open neighbourhood $W_x$ such $W_x\subseteq W$. So $W$ is open.

Now let us answer the question:

Take $y \in Y$. Let $A$ be an open subset of $Y$ such that $y\in A$. Let $C$ be the component of $A$ such that $y\in C$. In order to prove that $C$ in an open connected neighbourhood of $y$ in $Y$, it is enough to prove $C$ is open.

Proof that $C$ is open:

Since the topology in Y is the quotient topology, we have that $f^{-1}(A)$ is open and that $f^{-1}(C)$ is a union of components of $f^{-1}(A)$. Since $X$ is locally connected, the components of open sets in X are open. So $f^{-1}(C)$ is a union of open sets in X. So $f^{-1}(C)$ is open. So $C$ is open in Y

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I'm sure you mean locally connected in the first sentence :-) $\endgroup$ – Stefan Hamcke Jul 21 '15 at 19:36
  • $\begingroup$ Also, one should require that the connected neighborhood $W_x$ is in $U$, since $W$ contains all connected neighborhoods of $x$ which are subsets of $U$. $\endgroup$ – Stefan Hamcke Jul 21 '15 at 19:40
  • $\begingroup$ @stefan Thanks. $\endgroup$ – Ramiro Jul 21 '15 at 20:14

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.