If $f:X\longrightarrow Y$ is surjective, closed and continuous then $Y$ is weakly locally connected provided $X$ is it.

Question

First of all let's we give the following definition.

Definition

A space $X$ is weakly locally connected if for each neighborhood $N_x$ of any $x\in X$ there exist a connected subspace $F_x$ and a neighborhood $U_x$ such that $$ U_x\subseteq F_x\subseteq N_x $$

So I am trying to understand if $f$ is a surjective, closed and continuous map from a space $X$ to a space $Y$ then this last is weakly locally connected whether the first is it.

So by continuity $f^{-1}[V]$ is open if $V$ is an open neighborhood of $y\in Y$ and thus by surjectivity for any $x\in f^{-1}[V]$ there exists a connected set $F_x$ and an open set $U_x$ such that $$ x\in U_x\subseteq F_x\subseteq f^{-1}[V] $$ So putting $$ F:=\bigcup_{x\in f^{-1}[y]}F_x $$ we observe that $$ U:=\bigcup_{x\in f^{-1}[y]}U_x $$ is an open set such that $$ f^{-1}[y]\subseteq U\subseteq F\subseteq f^{-1}[V] $$ so that by closedness $Y\setminus f[X\setminus U]$ is an open set such that $$ \tag{1}\label{1}y\in Y\setminus f[X\setminus U]\subseteq f[U]\subseteq f[F]\subseteq V $$ Now we observe that $y$ is in $f[F_x]$ for any $x\in f^{-1}[y]$ so that the intersection $$ \bigcap_{x\in f^{-1}[y]} f[F_x] $$ is not empty and thus the union $$ \bigcup_{x\in f^{-1}[y]}f[F_x] $$ is connected because by continuity $f[F_x]$ is connected for any $x\in f^{-1}[y]$. So by $\eqref{1}$ we infer that $Y$ is weakly locally connected.

So I ask if what I proved is true and thus if it is then I ask if I well proved it whereas I ask to disprove it with a counterexample. So could someone help me, please?

Answer 1

The proof is correct and actually it shows that if $f$ is a closed and continuous surjection from $X$ to $Y$ then $Y$ is locally connected provide $X$ is it because weakly locally connection is just locally connection has showed to follow.

First of all we give the following definition

Definition

For a point $x$ of a topological space $X$ the component $C_x$ of $X$ is the union of all connected subspace of $X$ containing it.

So immediately we observe that singletons are trivially connected so that $C_x$ is defined for any $x\in X$. Moreover we observe that if $C_1$ and $C_2$ are two components of $X$ with not empty intersection then their union $C_1\cup C_2$ would be a component of any $x_0\in C_1\cap C_2$ but this is absurd because $C_1$ or $C_2$ would not be the union of all connected subspace containing $x_0\in X$: so we conclude that the collection of connected components is a partition. Now we give the following definition.

Definition

A space $X$ is locally connected if any $x\in X$ has a local base $\mathcal B(x)$ of connected neighborhoods.

So let's we prove the following result bearing in mind that a subset $H$ of any subspace $Y$ of a space $X$ is connected with repsect $Y$ if and only if it is connected with respect $X$.

Theorem

For any topological space $X$ the following statements are equivalent:

1. $X$ is locally connected;
2. the component of the open sets are open;
3. for each neighborhood $N_x$ of any $x\in X$ there exist a connected subspace $H_x$ and an open set $A_x$ such that $$x\in A_x\subseteq H_x\subseteq N_x$$

Proof.

1. If $X$ is locally connected then any $x\in A$ with $A$ open has a connected neighborhood $N_x$ there contained. So the component $C_x$ of $x$ in $A$ is the greatest connected subspace of $A$ containing $x$ so that if $N_x$ is connected with respect $X$ and if it is contained in $A$ then the inclusion $$ N_x\subseteq C_x $$ must holds. Now we know that if $x_0\in C_x$ then surely the equality $$ C_x=C_{x_0} $$ holds so that by the above arguments we conclude there exists a neighborhood $N_0$ of $x_0$ contained in $C_x$ and thus finally we conclude that $C_x$ is open.
2. If $A(x)$ is an open neighborhood of any $x\in X$ then the component $C_{A(x)}$ of $x$ in $A(x)$ is connected in $X$ so that it is there open and thus $C_{A(x)}$ is an open connected neighborhood of $x$ such that $$ C_{A(x)}\subseteq A(x) $$ which proves that $3$ holds: indeed if $N(x)$ is a neighborhod of $x$ then there exists an open set $A(x)$ such that $$ x\in A(x)\subseteq N(x) $$ so that we can put $$ A_x,H_x:=C_{A(x)} $$
3. So we know that $A_x$ is an opne set containing $x$ and contained in $H_x$ so that $H_x$ is connected neighborhood of $x$ contained in $N_x$ and thus for any $N_0\in\mathcal N(x)$ the collection $$ \mathcal C(N_0):=\{N_x\in\mathcal N(x):N_x\,\text{contained in $N_0$ and connected}\} $$ is not empty and thus finally che collection $$ \mathcal C(x):=\bigcup_{N_x\in\mathcal N(x)}\mathcal C(N_x) $$ is a local connected base for $x$.

However, we observe that if $V\in\mathcal P(Y)$ is such that $f^{-1}[V]$ is open in $X$ then its complement is closed but actually the equality $$ f^{-1}[Y\setminus V]=f^{-1}[Y]\setminus f^{-1}[V]=X\setminus f^{-1}[V] $$ holds and thus by closedness and by surjectivity $Y\setminus V$ is closed, that is $V$ is open which means that $f$ is a final map or rather that $Y$ has the final topology induced by $f$: so actually the result can be proved by this more general theorem proved by Paul Frost.