Show that $CX$ is locally (path-)connected iff $X$ is

Question

I'm doing some excercise in Lee's book. I've come up with a solution but not really sure. I also want to see other people solutions too. Here is the problem:

Let $X$ be any topological space and $CX$ is the cone $(X \times I)/(X \times \{1\})$. Show that $CX$ is locally (path-)connected $\Leftrightarrow$ $X$ is.

Note : I'm aware of the answer here that use the fact that quotient space of a locally (path-)connected space is locally (path-)connected. But since in Lee's book does not prove this fact (it's even not in the problem), i think i should solve this problem by constructing explicit connected basis for $CX$ and $X$. Forgive me for my silly reason.

I think i can handle the $\Rightarrow$ part. But i stuck for $\Leftarrow$ part. Can anyone suggest a hint for me to do this ? Or any solution without construct a basis for $CX$ ? Thank you.

$\textbf{Edit :}$

I know that if i can construct a neighbourhood basis for vertex of $CX$, then the problem solved. I have two ideas:

1). First, suppose $(B_{\alpha})_{\alpha \in A}$ is the connected basis for $X$, then for each $\alpha$, define the set $$ T_{\alpha} = \{ B_{\alpha} \times (t,1] \mid \text{ for all } 0\leq t < 1\} $$ the elements are connected open subset of $X \times I$. Now, take (not empty) one element for each $T_\alpha$, say $B_{\alpha} \times (t_{\alpha},1]$, and then take their union $\bigcup_{\alpha \in A} B_{\alpha} \times (t_{\alpha},1]$. Then define a set $\mathcal{U}$ as the collection of all of these possible union. Each of these union is an open subset in $X \times I$ contain $X \times \{0\}$. The image of these collection under quotient map $q$, denoted by $\mathcal{B}$, is then a collection of connected neighbourhoods of the vertex.

is $\mathcal{B}$ is a well-defined neighbourhood basis for the vertex ?

2). I have shown that $CX$ is contractible and path-connected. But how can i use this fact to construct a neighbourhood basis for the vertex ?

I hope someone would help me with this. I've stuck here for days (felt so embarrased). I'am self studying and don't have anyone here to talk about topology. Thank you.

Answer 1

Hint:

To show $\Leftarrow$ analyze what happens (looking at open basis sets) when you remove the apex $X \times \{1\}$ from the cone. Can you get results for the topological product $X \times [0,1)$ (which is embedded in $CX$)?

Next, show that there are basis open sets about the apex of $CX$ that are path-connected (you get an obvious path connecting each point up to the apex). This is always true and does not depend on properties of $X$.

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For every open in $CX$ containing the apex we can find a smaller open set containing the apex defined by using open rectangles on the product topology from $P = X \times [0,1]$. If $X$ is locally path-connected we can do more:

Every open subset of $P$ containing $X \times \{1\}$ contains an open set of the form

$\quad \bigcup_{x \in X} \; (\,U_x \times (\alpha_x, 1]\,)$

where $U_x$ is open and path-connected in $X$ with $x \in U_x$ and $\alpha_x \lt 1$, so that $U_x \times (\alpha_x, 1]$ is path-connected.

Note that when $X$ is compact we find a 'nicer more uniform' basis about the apex in $CX$:

$\quad (\,X \times (1 - \frac{1}{n}, 1]\,) \;/\; (\,X \times \{1\}\,)$

Answer 2

(1) If $CX=I\times X/(0,x)\sim (0,x_0)$ is locally connected, then note that $U:=(1-\epsilon,1]\times X$ is open in $CX$. Note that topology in $U$ is corresponded to that of $X$. So since $U$ is locally connected, then so is $X$.

(2) Assume that $X$ is locally connected. Then $p$ is not a vertex point in $CX$. Then there is open $U'$ containing $p$ s.t. $U'$ is not locally connected.

There are two types : $U'=(a,b)\times A$ where $A$ is open in $X$ or $U'=(a,1]\times A$.

If $U'$ has a sepration, then so does $A$.

Open set in $CX$ not considered is $V$ at a vertex point. Clearly, $V$ is path connected.