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

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...

• Actually that set is open. It's because of a particular property of quotient maps (such as $f$). Do you know this property? – john Jul 21 '15 at 17:33
• 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? – amirbd89 Jul 21 '15 at 17:36
• 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. – john Jul 21 '15 at 18:50
• 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 $X$. – john Jul 21 '15 at 18:53

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

• I'm sure you mean locally connected in the first sentence :-) – Stefan Hamcke Jul 21 '15 at 19:36
• 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 $U$. – Stefan Hamcke Jul 21 '15 at 19:40
• @stefan Thanks. – Ramiro Jul 21 '15 at 20:14