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\begin{document}

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\section*{locally connected topological space}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{Definition}{Definition}\dotfill \pageref*{Definition} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{the_category_of_locally_connected_spaces}{The category of locally connected spaces}\dotfill \pageref*{the_category_of_locally_connected_spaces} \linebreak
\noindent\hyperlink{cohesion_over_sets}{Cohesion over sets}\dotfill \pageref*{cohesion_over_sets} \linebreak
\noindent\hyperlink{quotients_of_locally_connected_spaces}{Quotients of locally connected spaces}\dotfill \pageref*{quotients_of_locally_connected_spaces} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

It is not generally true that a [[topological space]] is the [[disjoint union space]] ([[coproduct]] in [[Top]]) of its [[connected components]]. The spaces such that this is true for all open subspaces are the \emph{locally connected topological spaces}.

\hypertarget{Definition}{}\subsection*{{Definition}}\label{Definition}

\begin{defn}
\label{LocallyConnected}\hypertarget{LocallyConnected}{}
\textbf{(locally connected topological space)}

A [[topological space]] $X$ is \textbf{locally connected} if every point has a [[neighborhood basis]] of [[connected topological space|connected]] [[open subsets]].

\end{defn}
\begin{prop}
\label{AlternativeCharacterizationsOfLocalConnectivity}\hypertarget{AlternativeCharacterizationsOfLocalConnectivity}{}
\textbf{(alternative characterizations of local connectivity)}

For $X$ a [[topological space]], then the following are equivalent:

\begin{enumerate}%
\item $X$ is locally connected (def. \ref{LocallyConnected});


\item every [[connected component]] of every open [[subspace]] of $X$ is open;


\item every [[open subset]], as a [[topological subspace]], is the [[disjoint union space]] ([[coproduct]] in [[Top]]) of its [[connected components]].



\end{enumerate}
In particular, in a locally connected space, every connected component $S$ is a [[clopen subset]]; hence connected components and quasi-components coincide.

\end{prop}
\begin{proof}
$\,$

1) $\Rightarrow$ 2)

Assume $X$ is locally connected, and let $U \subset X$ be an open subset with $U_0 \subset U$ a connected component. We need to show that $U_0$ is open.

Consider any point $x \in U_0$. Since then also $x \in U$, the defintion of local connectedness, def. \ref{LocallyConnected}, implies that there is a connected open neighbourhood $U_{x,0}$ of $X$. Observe that this must be contained in $U_0$, for if it were not then $U_0 \cup U_{x,0}$ were a larger open connected open neighbourhood, contradicting the maximality of the connected component $U_0$.

Hence $U_0 = \underset{x \in U_0}{\cup} U_{x,0}$ is a union of open subsets, and hence itself open.

2) $\Rightarrow$ 3)

Now assume that every connected component of every open subset $U$ is open. Since the connected components generally consitute a [[cover]] of $X$ by [[disjoint subsets]] this means that now they for an [[open cover]] by disjoint subsets. But by forming intersections with the cover this implies that every open subset of $U$ is the disjoint union of open subsets of the connected components (and of course every union of open subsets of the connected components is still open in $U$), which is the definition of the topology on the [[disjoint union space]] of the connected components.

3) $\Rightarrow$ 1)

Finally assume that every open subspace is the disjoint union of its connected components. Let $x$ be a point and $U_x \supset \{x\}$ a neighbourhood. We need to show that $U_x$ contains a connected neighbourhood of $x$.

But, by definition, $U_x$ contains an open neighbourhood of $x$ and by assumption this decomposes as the disjoint union of its connected components. One of these contains $x$. Since in a [[disjoint union space]] all summands are open, this is the required connected open neighbourhod.

\end{proof}
\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{example}
\label{LocallyConnectedEuclideanSpace}\hypertarget{LocallyConnectedEuclideanSpace}{}
\textbf{([[Euclidean space]] is locally connected)}

For $n \in \mathbb{N}$ the [[Euclidean space]] $\mathbb{R}^n$ (with its [[metric topology]]) is locally connected.

\end{example}
\begin{proof}
By nature of the Euclidean metric topology, every neighbourhood $U_x$ of a point $x$ contains an [[open ball]] containing $x$. Moreover, every open ball clearly contains an open cube, hence a [[product space]] $\underset{i \in \{1, \cdots, n\}}{\prod} (x_i-\epsilon, x_i + \epsilon)$ of [[open intervals]] which is still a neighbourhood of $x$.

Now intervals are connected (by \href{connected+space#ConnectedSubspacesOfRealLineAreTheIntervals}{this example}) and products of connected spaces are connected (by \href{connected+space#ProductSpaceOfConnectedSpacesIsConnected}{this example}). This shows that ever open neighbourhood contains a connected neighbourhood.

\end{proof}
\begin{prop}
\label{LocallyConnectedOpenSubspaceOfLocallyConnectedSpace}\hypertarget{LocallyConnectedOpenSubspaceOfLocallyConnectedSpace}{}
\textbf{([[open subset|open]] [[subspace]] of locally connected space is locally connected)}

Every [[open subset|open]] [[subspace]] of a locally connected space is itself locally connected

\end{prop}
\begin{proof}
This is immediate from def. \ref{LocallyConnected}.

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
\textbf{Warning}

A [[connected topological space]] need not be locally connected.

\end{remark}
\begin{example}
\label{}\hypertarget{}{}
The [[topologist's sine curve]] is connected but not locally connected.

\end{example}
Examples of locally connected spaces include [[manifold|topological manifolds]].

Finally,

\begin{defn}
\label{}\hypertarget{}{}
A space $X$ is \textbf{[[totally disconnected topological space]]} if its [[connected components]] are precisely the [[singletons]] of $X$.

\end{defn}
In other words, a space is totally disconnected if its coreflection into $LocConn$ is discrete. Such spaces recur in the study of [[Stone spaces]].

The category of totally disconnected spaces is a reflective subcategory of $Top$. The reflector sends a space $X$ to the space $X/\sim$ whose points are the connected components of $X$, endowed with the quotient topology induced by the projection $q: X \to X/\sim$. Details may be found at [[totally disconnected space]].

\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\hypertarget{the_category_of_locally_connected_spaces}{}\subsubsection*{{The category of locally connected spaces}}\label{the_category_of_locally_connected_spaces}

Let $i \colon LocConn \hookrightarrow Top$ be the [[full subcategory]] inclusion of locally connected spaces into all of [[Top]]. The following result is straightforward but useful.

\begin{theorem}
\label{coref}\hypertarget{coref}{}
$LocConn$ is a [[coreflective subcategory]] of $Top$, i.e., the inclusion $i$ has a [[right adjoint]] $R$. For $X$ a given space, $R(X)$ has the same underlying set as $X$, topologized by letting connected components of open subspaces of $X$ generate a topology.

\end{theorem}
Being a coreflective category of a complete and cocomplete category, the category $LocConn$ is also complete and cocomplete. Of course, limits and particularly \emph{infinite} products in $LocConn$ are not calculated as they are in $Top$; rather one takes the limit in $Top$ and \emph{then} retopologizes it according to Theorem \ref{coref}. (For \emph{finite} products of locally connected spaces, we can just take the product in $Top$ -- the result will be again locally connected.)

\hypertarget{cohesion_over_sets}{}\subsubsection*{{Cohesion over sets}}\label{cohesion_over_sets}

Let $\Gamma \colon LocConn \to Set$ be the underlying set functor, and let $\nabla, \Delta \colon Set \to LocConn$ be the functors which assign to a set the same set equipped with the codiscrete and discrete topologies, respectively. Let $\Pi_0 \colon LocConn \to Set$ be the functor which assigns to a locally connected space the set of its connected components.

\begin{theorem}
\label{}\hypertarget{}{}
There is an [[adjoint quadruple]] of [[adjoint functors]]

\begin{displaymath}
\Pi_0 \dashv \Delta \dashv \Gamma \dashv \nabla \colon Set \to LocConn
\end{displaymath}
and moreover, the functor $\Pi_0$ preserves finite products.

\end{theorem}
The proof is largely straightforward; we point out that the continuity of the unit $X \to \Delta \Pi_0 X$ is immediate from a locally connected space's being the coproduct of its connected components. As for $\Pi_0$ preserving finite products, write locally connected spaces $X$, $Y$ as coproducts of connected spaces

\begin{displaymath}
X = \sum_i C_i; \qquad Y = \sum_j D_j;
\end{displaymath}
then their product in $LocConn$ coincides with their product in $Top$, and is

\begin{displaymath}
X \times Y \cong \sum_{i, j} C_i \times D_j
\end{displaymath}
where each summand $C_i \times D_j$ is connected by Result \ref{3}. From this it is immediate that $\Pi_0$ preserves finite products.

Accordingly the [[category of sheaves]] on a locally connected space is a [[locally connected topos]]. For related discussions, see also [[cohesive topos]].

\hypertarget{quotients_of_locally_connected_spaces}{}\subsubsection*{{Quotients of locally connected spaces}}\label{quotients_of_locally_connected_spaces}

\begin{lemma}
\label{quot}\hypertarget{quot}{}
A [[quotient space]] of a locally connected space $X$ is also locally connected.

\end{lemma}
\begin{proof}
Suppose $q: X \to Y$ is a quotient map, and let $V \subseteq Y$ be an open neighborhood of $y \in Y$. Let $C(y)$ be the connected component of $y$ in $V$; we must show $C(y)$ is open in $Y$. For that it suffices that $C = q^{-1}(C(y))$ be open in $X$, or that each $x \in C$ is an interior point. Since $X$ is locally connected, the connected component $U_x$ of $x$ in $q^{-1}(V)$ is open, and the subset $q(U_x) \subseteq V$ is connected, and therefore $q(U_x) \subseteq C(y)$ (as $C(y)$ is the maximal connected subset of $V$ containing $q(x)$). Hence $U_x \subseteq q^{-1}(C(y)) = C$, proving that $x$ is interior to $C$, as desired.

\end{proof}
The conclusion does not follow if $q: X \to Y$ is merely surjective; e.g., there is a surjective (continuous) map from $\mathbb{R}$ to (a version of) the [[Warsaw circle]], but the latter is not locally connected.

\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[totally disconnected topological space]]


\item [[locally connected topos]]



\end{itemize}
[[!redirects locally connected space]] [[!redirects locally connected spaces]] [[!redirects locally connected topological space]] [[!redirects locally connected topological spaces]]



\end{document}