Local connectedness is preserved under retractions I want to show that if $X$ is a locally connected topological space, $A\subseteq X$ is a subspace and $f:X \rightarrow A$ is continuous such that $f|_{A} = Id_{A}$, then $A$ must be locally connected as well.
My progress so far:
Take $U\subseteq A$ $A-$open and $x\in U$. Since $f^{-1} (U)$ is $X-$open and $x\in f^{-1} (U)$, there exists a connected, $X-$open subset $V\subseteq X$ such that $x\in V\subseteq f^{-1}(U)$.
Now we have $x\in V\cap A \subseteq f^{-1}(U)\cap A = U$
My guess is that $V\cap A$ should be connected, but I am unsure if this is correct.
Any help would be appreciated!
 A: It is easier to instead show that components of open sets are open, which is equivalent to local connectedness. Let $r\colon X\to A$ be the retraction and suppose $N$ is an open set in $A$ and $C$ is a component of $N$.
Let $p\in N$ and $U = r^{-1}(N)$. By continuity, $U$ is open and by local connectedness, there is a connected open $V$ such that $p\in V\subseteq U$. Again by continuity, $r(V)$ is a connected set containing $p$, hence $r(V)\subseteq C$.
Now, $p\in V\cap A\subseteq r(V)$ and $V\cap A$ is open in $A$, therefore, $p\in V\cap A\subseteq C$, hence $C$ is an open set.
Therefore, the connected components of $N$ are open, hence $A$ is locally connected.
A: Let K be the nonnegative x and y axis.
Project every point of the 1st quadrant at a 45 degree angle o  K.
Thus K is a retract of the first quadrant.
In general, every V is retract of the space between the two lines.  
Let C be the comb space of all lines from (0,0) to (1,1/n)
for all n in N and the line from (0,0) to (1,0).  
In view of the retracts of all the V's of K can the first quadrant be retracted to K?  Thus be a counterexample.
