$X$ 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...
 A: 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
