When is the mapping telescope homotopy equivalent to the space? Let $(X,A)$ be a relative CW-complex with filtration $A = X_{-1} \subseteq X_0 \subseteq X_1 \subseteq ... X = \operatorname*{colim} \limits_{n \in \mathbb N_0} X_n$. It is a well known result (see for example Hatcher p.138) that $X$ is homotopy equivalent to the mapping telescope of the filtration, which is the essential step in showing $H_\bullet(X,A) \cong \operatorname{colim} H_\bullet(X_n,A)$ for any generalized homology theory $(H_\bullet, \partial_\bullet)$. Is this specific to CW-complexes? In other words

Under which circumstances is the mapping cone of a topological space $X$ with a filtration $(X_n)_n$ homotopy equivalent to the mapping telescope of the filtration?

Adapting the proof of the CW-case it seems reasonable to me to suppose that $X_n$ is a closed neighborhood deformation retract of $X_{n+1}$. Yet, in the construction of the mapping telescope we use spaces of the form $[0,1]\times X_n$, so this naturally leads to the question

If $X$ is a closed neighborhood deformation retract of $Y$, does it hold that $X\times[0,1]$ is a closed neighborhood deformation retract of $Y \times [0,1]$?

I could not find a reference for a general result. I would really appreciate any help.
 A: If $i:A \rightarrow X$ is a closed neighborhood deformation retract, there is an open neighborhood $U$ of $A$ in $X$ such that the restriction $i:A \rightarrow U$ is a deformation retract. This means there is a retraction $r: U \rightarrow A$ with $ri = 1_A$ and a homotopy $h: I \times U \rightarrow U$, which satisfies
$$\begin{array}{rcl}
h(0,u) &=& ir(u)\\
h(1,u) &=& u\\
h(t,a) &=& a,
\end{array}$$
using suggestive notation. We want to show that $I \times i: I \times A \rightarrow I \times U$ is a deformation retract. Clearly $I \times r: I \times U \rightarrow I \times A$ is a retraction. Furthermore, given the transposition $$\begin{array}{rcl}
\tau:I \times I &\rightarrow& I \times I\\ (t,s) &\mapsto& (s,t),
\end{array}$$ the composition $$k: I \times I \times U \overset{\tau \times U}{\longrightarrow} I \times I \times U \overset{I \times h}{\longrightarrow} I \times U$$ is continuous and satisfies
$$\begin{array}{rcl}
k(0,s,u) &=& (s,h(0,u)) = (s,ir(u))\\
k(1,s,u) &=& (s,h(1,u)) = (s,u)\\
k(t,s,a) &=& (s,h(t,a)) = (s,a),
\end{array}$$
which turns $I \times i: I \times A \rightarrow I \times U$ into a deformation retract.
Remark I hope I did not make a mistake. In case I did, I won't accept this answer until I have found someone who verified this.
PS As @William pointed out in the comments, the theory of model categories generalizes problems and arguments like this. But because I am currently attending my first lecture on algebraic topology, I will have to come back to this at a later point.
