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?