Neighborhood deformation retracts vs cofibrations I got really confused over the different notions of neighborhood deformation retracts and cofibrations one can find in various sources on algebraic topology and alike, so I would really appreciate, if someone could help me out. I did not find a question immediately linking the various notions, so I hope this is not a duplicate.
I assume $A \subseteq X$ to be a closed subspace. How do the following definitions correlate?

Definition 1
$A$ is a strong neighborhood deformation retract of $X$, if there is an open neighborhood $A \subseteq N \subseteq X$, such that the inclusion $i:A \rightarrow X$ has a retraction $r:N \rightarrow A$ with $ri = 1_A$ and $ir \sim 1_N$ via a homotopy $H:N\times[0,1] \rightarrow N$ satisfying $H(a,t) = a$ for $a\in A$.
Definition 2
$A$ is a neighborhood deformation retract of $X$, if there is an open neighborhood $A \subseteq N \subseteq X$, such that the inclusion $i:A \rightarrow X$ has a retraction $r:N \rightarrow A$ with $ri = 1_A$ for which there is a homotopy $H:N\times[0,1] \rightarrow X$ satisfying $H(x,0)=x$, $H(x,1)\in A$, and $H(a,t) = a$ for $x\in N$, $a\in A$, and $t\in I$.
Definition 3
$(X,A)$ is a NDR-pair (ncatlab) or $A$ is a neighborhood deformation retract of $X$ (wikipedia), if there are maps $h:I\times X \rightarrow X$ and $u:X \rightarrow I$, which satisfy

*

*$h(a,t) = a$

*$h(x,0) = x$

*$u^{-1}(\{0\}) = A$

*$h(x,t)\in A$ if $u(x)<t$.

ncatlab section 3 or wikipedia, cofibrations and NDRs

Ncatlab mentions that the canonical inclusion $i:A \rightarrow X$ has a homotopy left inverse, if and only if it has a retraction $r:X \rightarrow A$ (ie. $ri = 1_A$). This remark confuses me, as in my understanding this would make $A$ a deformation retract of $X$ instead of a neighborhood deformation retract.
Wikipedia mentions at the same place as definition 4 that it is equivalent to the following definition of cofibration.

Definition 4
The inclusion $i:A \rightarrow X$ is a cofibration, if it has the homotopy extension property, ie. for any morphism $f:A\times I \cup X\times \{1\}\rightarrow T$ there exists a (not necessarily unique) extension $\tilde{f}:I \times X \rightarrow T$ along the inclusion $j:A\times I \cup X\times \{0\} \rightarrow X\times I$, meaning that $f = \tilde{f}i$. wikipedia homotopy extension property

According to Groth Prop. 3 being a cofibration is equivalent to $j:A\times I\cup X\times\{0\} \rightarrow X\times I$ having a retraction.
My initial goal was to show that, given a closed neighborhood deformation retract $i:A \rightarrow X$, the map $j:A\times I\cup X\times\{0\} \rightarrow I \times X$ is a deformation retract. Instead, I managed to confuse myself to an extent, which made it impossible for me to find relations between the four definitions given here, yet alone to approach my initial problem. I really hope someone can help me out. Regardless, huge thanks to anyone who read up until here!
 A: We start by proving
$(4)\Rightarrow(3)$ Taking $f$ as the identity we get a retraction $r:X\times I\rightarrow A\times I\cup X\times0$. Fixing one such we set $u:X\rightarrow I$ to be the map
$$u(x)=\sup_{t\in I}|t-pr_2\circ r(x,0)|,\qquad x\in X.$$
Also let $h:X\times I\rightarrow X$ be the homotopy
$$h(x,t)=pr_1\circ r(x,t),\qquad t\in I,x\in X.$$
Then all required properties are immediate. $\blacksquare$
$(3)\Rightarrow(4)$ We have the maps $u,h$ and need to define a retraction $r$ to the inclusion $A\times I\cup X\times\{0\}\subseteq X\times I$. This is given by
$$r(x,t)=\begin{cases}(h(t,x),0)&t\leq u(x)\\
(h(t,x),t-u(x))& t\geq u(x)\end{cases}$$
You check easily that it is well-defined. Given $f:A\times I\cup X\times 0\rightarrow T$ the extension is now $\widetilde f=fr:X\times I\rightarrow T$. $\blacksquare$
Thus $(3)$ and $(4)$ are equivalent and imply that the inclusion of the closed subspace $A\subseteq X$ is a cofibration.
$(3)\Rightarrow(2)$ Put $N=u^{-1}[0,1)$ and let $H=h|_{N\times I}$. $\blacksquare$
The last implication is not reversible in general. It turns out the presence of the function $u$ is extremely important. However, we can go backwards if we assume the additional condition
$(\ast)$: There is a map $v:X\rightarrow I$ such that $A=v^{-1}(0)$ and $N=v^{-1}[0,1)$.
Evidently $(3)\Rightarrow(2)+(\ast)$.
$(2)+(\ast)\Rightarrow(3)$ Define a retraction $r:X\times I\rightarrow A\times I\cup X\times 0$ by
$$r(x,t)=\begin{cases}
(x,t)&x\in v^{-1}(0)\\
(h(x,t/2v(x)),0)&x\in v^{-1}(0,1/2]\;\text{and}\;t\leq2v(x)\\
(h(x,1),t+2v(x))&x\in v^{-1}(0,1/2]\;\text{and}\;2v(x)\leq t<1\\
(h(x,2(1-v(x))t),0)&x\in v^{-1}[1/2,1)\\
(x,0)&x\in v^{-1}(1).\quad\blacksquare
\end{cases}$$
At this stage we have shown that $(2)+(\ast)\Leftrightarrow(3)\Leftrightarrow(4)$ are all equivalent. Note that sufficient conditions for $(\ast)$ to hold are given by any of the following.

*

*$X$ is perfectly normal and $A\subseteq X$ is closed.

*$X$ is normal and $A\subseteq X$ is a closed $G_\delta$-set.

*$X$ is Tychonoff and $A\subseteq X$ is a compact $G_\delta$-set.

Thus $(X,A)$ having any of these properties is sufficient for $(3)\Rightarrow(4)$ with no apriori knowledge of $v$. Note that every metric space (so every manifold) and every CW complex is perfectly normal. A $G_\delta$-set is a subset which is the intersection of countably many open sets.
Now, $(3)\Rightarrow(2)$ and obviously $(1)\Rightarrow(2)$, with neither implication reversible in general. Unfortunately there are also no direct implications between $(1)$ and $(3)$, as we have counterexamples to the contrary.
As for your last question, if $(X,A)$ is a closed NDR pair (def. 3), then we have a retraction $r:X\times I\rightarrow A\times I\cup X\times 0$, and a homotopy
$$H_s(x,t)=(pr_1\circ r(x,st),(1-s)t+s pr_2\circ r(x,t))$$
Thus the inclusion of $A\times I\cup X\times 0$ into the cylinder is a strong deformation retraction in this case.
