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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.

Alex Ortiz
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D. Hershko
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2 Answers2

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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.

Henno Brandsma
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  • 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 – jasmine May 29 '21 at 14:43
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    @jasmine I start with an arbitrary point of $C$. That’s why I can conclude for all points. – Henno Brandsma May 29 '21 at 15:10
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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.

mathieu
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