# Local Path Connectedness and Cones

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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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.
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).