If $A$ has a neighbourhood in $X$ that deformation retracts onto $A$ then does $(X,A)$ have HEP? Let $(X,A)$ be a topological pair. Assume that $A$ has a neighbourhood $V$ such that $V$ deformation retracts onto $A$. Then can we say that $(X,A)$ has the homotopy extension property?
Edit: I know this is true if $A$ has a mapping cylinder neighbourhood i.e. a closed nbhd $N$ of $A$ such that there exixts a map $f: B\rightarrow A$ where $B=N-int(N)$ and $N \cong Cyl(f)$ with $ B\rightarrow N$ corresponds to the inclusion at the top of the mapping cylinder. 

This does not hold. Take for example $X=I\times \{0\}\cup I\times \{\frac{1}{n}|n\in\mathbb{Z}_+\}\cup \{0\}\times I$, and take $A=\{(0,1)\}$. 
The inclusion of $A$ into $X$ is not a cofibration. To see this take $Y$ to be $Y=X\times \{0\}\cup A\times I$ and $f:X\times \{0\}\cup A\times I\to Y$ the identity map. Extending this homotopy is then equivalent to finding a retraction of the subspace $Y\subseteq X\times I$. No such retraction exists.
On the other hand, $X$ is contractable.
