Let $X$ be a locally connected space and $f:X \to Y$ a continuous closed surjection. Show that $Y$ is locally connected. 
Let $X$ be a locally connected space and $f:X  \to Y$ a continuous closed surjection. Show that $Y$ is locally connected.

Let $V \subset Y$ be an open set and $C$ a component containing $V$. To prove that $Y$ is locally connected we need to show that $C$ is open.
Since $f$ is continuous $f^{-1}(V)$ is open and $f^{-1}(C)$ is closed.
If I know somehow could show that $f^{-1}(C)$ is also open, then since $f$ is a continuous closed surjection it's a quotient map and $C$ would be open. How can we find some data to show that $f^{-1}(C)$ is open? I think we need to use local connectivity of $X$ somehow but that only gives me that each of the components of $X$ are also open.
 A: We shall give a proof both for local path connectedness and for local connectedness. Note that continuous closed surjections are quotient maps.
Theorem. Let $f : X \to Y$ be a quotient map (which is a surjective map such that $V \subset Y$ is open in $Y$ iff $f^{-1}(U) \subset X$ is open in $X$) and let $X$ be locally (path) connected. Then $Y$ is locally (path) connected.
Proof. Let $y \in Y$ and $V$ be an open neighborhood of $y$ in $Y$. Let $C$ be the (path) component of $V$ containing $y$. We shall show that $f^{-1}(C)$ is open in $X$ which implies that $C$ is open in $Y$.
This proves that $Y$ has a basis of (path) connected open sets.
So let $x \in f^{-1}(C)$. Then $f(x) \in f(f^{-1}(C)) = C \subset V$, i.e. $x \in f^{-1}(V)$. Since $f^{-1}(V)$ is open in $X$, we find a (path) connected open neigborhood $U_x$ of $x$ in $X$ such that $U_x \subset f^{-1}(V)$. The image $f(U_x)$ is (path) connected and we have $f(x) \in f(U_x) \subset f(f^{-1}(V)) = V$. Since $f(x) \in C$, we see that $C$ is the (path) component of $V$ containing $f(x)$, and since $f(U_x)$ is a (path) connected subset of $V$ containing $f(x)$, we conclude that $f(U_x) \subset C$. Therefore $U_x \subset f^{-1}(f(U_x)) \subset f^{-1}(C)$.
We have proved that each $x \in f^{-1}(C)$ has an open neighborhood $U_x$ in $X$ such that $U_x \subset f^{-1}(C)$; this means that $f^{-1}(C)$ is open in $X$.
Remark:
Each map $f : X \to Y$ which has a right inverse is a quotient map (a right inverse is a map $i : Y \to X$ such that $f \circ i = id_Y$).
In fact, if $V \subset Y$ is open in $Y$, then $f^{-1}(V) \subset X$ is open in $X$ by continuity of $f$. Conversely, for $V \subset Y$ such that $f^{-1}(V)$ is open in $X$, we see that $V = id_Y^{-1}(V) =  (f \circ i)^{-1}(V) = i^{-1}(f^{-1}(V))$ is open in $Y$ by continuity of $i$.
Note that each retraction $f : X \to Y$ has this property ( a retraction is a map from a space $X$ to a subspace $Y \subset X$ such that $f \mid_Y = id_Y$).  As a right inverse we can take the inclusion  map $i :Y \to X$.
We therefore get
Corollary. Each retract of a locally (path) connected space is locally (path) connected.
A: I don't know, what you mean with "component"? Path-connected, connected or quasi-connected component (latter of which are connected in locally connected spaces)? I am also not sure, why you need them: A space is locally connected at a point, if every neighborhood contains an open and connected neighborhood. Using the definition, I can offer the following sketch of a proof, which unfortnutly is still incomplete:
Since $f$ is a continuous and closed surjection, it is a quotient map as you mentioned already and therefore a subset $V\subset Y$ is open iff $f^{-1}(V)\subset X$ is open.
Let $y\in Y$ and $V\subset Y$ with $y\in V$ be a neighborhood. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of any point $x\in f^{-1}(y)\subset X$ and there exists at least one as $f$ is surjective. Since $X$ is locally connected, there is an open and connected neighborhood $U\subset f^{-1}(V)$ with $x\in U$. Since continuous images of connected spaces are connected, we know, that $f(U)\subset Y$ with $y=f(x)\in f(U)\subseteq V$ is connected.
The only thing left to show is, that either $f(U)$ is open (which holds if $f$ is bijective and therefore is a homeomorphism) or contains an open and connected neighborhood of $y$. Here I am stuck at the moment, but will continue to think about it.
