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I am trying to prove the following:

Let $X$ be a topological space. $CX$ is locally path connected if and only if $X$ is locally path connected, where $CX=X\times I/(X\times\{0\})$

I cannot seem to make much headway in either direction of the problem.

Assuming $CX$ is locally path connected means there exists a basis $\mathcal{B}$ such that every element of $\mathcal{B}$ is path connected. It's easy to see that $\mathcal{B}\cap X=\{X \cap B \: | \; B \in \mathcal{B}\}$ is a basis for $X$. I cannot quite see why the elements of $\mathcal{B}\cap X$ should be path connected. Let $\hat{B} \in\mathcal{B}\cap X$ and let $p,q \in \hat{B}$. $p$ and $q$ correspond with $(p,1)$ and $(q,1)$, respectively, in $B$ since $X\approx X\times\{1\}$. In $B$ there is a path $\alpha(t):[0,1] \rightarrow B$ such that $\alpha(0)=(p,1)$ and $\alpha(1)=(q,1)$. Perhaps the solution is obvious, but I cannot see how to turn a path in $B$ into a path in $\hat{B}$ or even if that should be possible. If $\alpha(t)$ was not entirely in $X \times \{1\}$, is projecting it onto $X\times \{1\}$ the solution?

I have also hit a wall in assuming $X$ is locally path connected. As such, there exists some path connected basis $\mathcal{C}$ of $X$. My first thought given $x,y \in CX$ was to find an open cone of the form $U\times I/(X\times \{0\})$, where $U$ is an open subset of $X$, such that $x,y \in U\times I/(X\times \{0\})$. I'm not even sure this is a basis of $CX$ and I cannot see why it should be.

Overall, I have no idea how to continue. Are any of my ideas leading in the right direction? If not, could anyone give me a recommendation in the right direction?

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up vote 2 down vote accepted

Here is a quick sketch of the argument. I can elaborate further if you wish.

Assume first that $X$ is locally path connected. Perhaps you know, and if not I encourage you to prove, that a finite product of locally (path) connected spaces is locally (path) connected and the image of a locally (path) connected space with a quotient map is locally (path) connected. Putting these two facts together shows that $CX$ must be locally path connected.

By the way, you are right in questioning whether your proposed basis really is one. In fact it is not (clearly you cannot get open sets that "don't touch the base of the cone" from your family).

For the converse, assume $CX$ is locally path connected. Your idea of projecting paths to the base of the cone is good, but has a slight problem. If the path contains the apex of the cone you obviously can't project it in a continuous way, so we have to fix this. I feel it is somewhat easier to work with another characterization of local path connectedness, i.e. that every point have a basis of path connected neighbourhoods. So take a point $x\in X$, which corresponds to $(x,1)\in CX$. Let $U$ be a neighbourhood of $x$ in $X$. Then (slightly abusing notation) $U\times (1/2,1]$ is a neighbourhood of $x$ in $CX$ and it doesn't contain the apex of the cone. By assumption, there exists a path connected neighbourhood $V$ of $x$ below $U\times(1/2,1]$. Projecting $V$ onto $X$ then gives us a path connected neighbourhood of $x$ in $X$ below $U$ (remember, we are basically in a product space now and projections are open maps).

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Thank you very much. – Holdsworth88 Aug 15 '12 at 19:03

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