# A quotient of a locally connected space is locally connected

Let $q : X \to Y$ be a quotient map, $X$ is a locally connected space. Show that $Y$ is also locally connected.

I will be thankful if some one could present a proof of this theorem, because I couldn't find one.

Thank you very much.

I'll use the following characterisation of local connectedness:

A space $X$ is locally connected iff for every open set $O$ of $X$, all connected components of $O$ are open in $X$.

This is a fact that is routinely taught about local connectedness and proofs can be found on this site.

Let's show that if $f: X \to Y$ is onto and quotient, and $X$ is locally connected, then $Y$ is locally connected.

Let $O$ be an open neighbourhood of a point $y \in Y$, and let $C_y$ be the component of $y$ in $O$. We want to show that $C_y$ is open, and so we need to show that $C= f^{-1}[C_y]$ is open: because $f$ is quotient we can then conclude that $C_y$ is open.

So let $x$ be any point in $C$. Then $f(x) \in C_y \subseteq O$, hence $x \in f^{-1}[O]$, which is open by continuity of $f$. So (using local connectedness of $X$) this $x$ has a connected neighbourhood $U_x$ such that $U_x \subseteq f^{-1}[O]$.

The set $f[U_x]$ is then also connected (as a continuous image of a connected set) and intersects $C_y$ in $f(x)$. So $C_y \cup f[U_x]$ is connected (and contains $y$) and is a subset of $O$, and as $C_y$ is a component of $O$ (so maximally connected inside $O$), and so $C_y \cup f[U_x] = C_y$ which implies that $f[U_x] \subseteq C_y$.

But recapping, the last equation just says that $U_x \subseteq f^{-1}[C_y] = C$ and so $x$ is an interior point of $C$.

So all points of $C$ are interior points and so $C$ is open. So, as we saw, $f$ being quotient then tells us $C_y$ is open, and by the characterisation, $Y$ is locally connected.

• Im not getting the last paragraph @Henno sir why you say all points of $C$ ? Here $C$ is connected component , so $C$ will be one-point sets $\implies C$ will contains only one point. So i think all points of $C$ will not correct Commented May 29, 2021 at 14:43
• @jasmine I start with an arbitrary point of $C$. That’s why I can conclude for all points. Commented May 29, 2021 at 15:10

Here is another way of expressing the same argument.

First a general fact, with no assumptions about $$X$$ and $$Y$$. If $$q:X\to Y$$ is continuous, then the pre-image of a connected component of $$Y$$ is a union of connected components of $$X$$. Indeed, if $$C$$ is a connected component of $$X$$, then $$q(C)$$ is connected because $$q$$ is continuous, so if $$D$$ is a connected component of $$Y$$ then $$q(C)$$ completely lies inside $$D$$ or outside $$D$$. In other words, $$C$$ completely lies inside $$q^{-1}(D)$$ or outside $$q^{-1}(D)$$.

Now, assuming that $$X$$ is locally connected, its connected components are open. By the previous fact, the pre-image of a connected component $$D$$ is a union of such sets so it is open. Assuming that $$q$$ is a quotient map, it implies that $$D$$ is open.

Finally, apply this argument to any open subspace $$O\subseteq Y$$, which is the quotient of the open subspace $$q^{-1}(O)\subseteq X$$, which is itself locally connected.