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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 a neighborhood $A \subseteq N \subseteq X$, such that the inclusion $i:A \rightarrow X$ has a retract $r:N \rightarrow A$ with $ri = 1_A$ and $ir \sim 1_N$ via a homotopy $h:[0,1]\times N \rightarrow N$ satisfying $h(t,a) = a$ for $a\in A$. Kammeyer Thm. 2.13

Definition 2
$A$ 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 retract $r:N \rightarrow A$ with $ri = 1_A$ and $ir \sim 1_N$ via a homotopy $h:[0,1]\times N \rightarrow N$ satisfying $h(t,a) = a$ for $a\in A$. In the proof of Thm. 2.13

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

  • $h(t,a) = a$
  • $h(1,x) = x$
  • $u^{-1}(\{0\}) = A$
  • $h(1,x)\in A$ if $u(x)<1$.

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 followig 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 $f:I \times A \cup \{1\} \times X \rightarrow T$ there exists a (not necessarily unique) extension $\tilde{f}:I \times X \rightarrow T$ along the inclusion $j:I \times A \cup \{1\} \times X \rightarrow I \times X$, meaning that $f = \tilde{f}i$. wikipedia homotopy extension property

According to Groth Prop. 3 being a cofibration is equivalent to $j:I \times A \cup \{1\} \times X \rightarrow I \times X$ having a retraction.

My initial goal was to show that, given a closed neighborhood deformation retract $i:A \rightarrow X$, the map $j:I \times A \cup \{1\} \times X \rightarrow I \times X$ 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!

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I tend to use the subspace $I\times A\cup 0\times X\subseteq I\times X$, as it tends to make the formulas easier to write down with the statement of definition 3.

$4)\Rightarrow 3)$ Taking $f$ as the identity we get a retraction $r:I\times X\rightarrow I\times A\cup0\times X$. Fixing one such we set $u:X\rightarrow I$ to be the map

$$u(x)=\sup_{t\in I}|t-pr_1\circ r(0,x)|,\qquad x\in X.$$

Also let $h:I\times X\rightarrow X$ be the homotopy

$$h(t,x)=pr_2\circ r(t,x),\qquad t\in I,x\in X.$$

Then all required properties are immediate. (Note that I corrected the last part of your statement of definition 3 to match your sources).

$3)\Rightarrow 4)$ We have the maps $u,h$ and need to define a retraction $r$ to the inclusion $A\times I\cup \{0\}\times X\subseteq I\times X$. This is given by

$$r(t,x)=\begin{cases}(0,h(t,x))&t\leq u(x)\\ (t-u(x),h(t,x))& t\geq u(x)\end{cases}$$

You check easily that it is well-defined. Given $f:A\times I\cup0\times X\rightarrow T$ the extension is now $\widetilde f=fr:X\times I\rightarrow T$.

Thus $3)$ and $4)$ are equivalent and imply that the inclusion of the closed subspace $A\subseteq X$ is a cofibration.

$3)\Rightarrow 2)\Rightarrow 1)$ Set $N=u^{-1}([0,1))$ and let $r:N\rightarrow A$ be the map $r(x)=h(u(x),x)$. The required homotopy $ir\simeq id_N$ is $(t,x)\mapsto h((1-t)u(x)+t,x)$.

Now the last implications are not reversible in general. It turns out the presence of the function $u$ is extremely important. If you have $u$, then you can go back, and Aguilar, Gitler and prieto give a proof under the additional assumption that $X$ is perfectly normal (pg. 94 of Algebraic Topology from a Homotopical Viewpoint).

As for your last question, if $(X,A)$ is a closed NDR pair (def. 3), then we have a retraction $r:I\times X\rightarrow I\times A\cup0\times X$, and a homotopy

$$H_s(t,x)=((1-s)t+s pr_1\circ r(t,x),pr_2\circ r(st,x))$$

Thus the inclusion of $I\times A\cup0\times X$ into the cylinder is a strong deformation retraction.

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  • $\begingroup$ Thank you so so much! This fills quite a hole in my current understanding of algebraic topology... $\endgroup$
    – PrudiiArca
    Feb 17 '20 at 16:59
  • $\begingroup$ In your proof 3) => 4), isn‘t $t > u(x)$ but $h(t, x) \notin A$ possible if for example $t < 1$? $\endgroup$
    – Bixxli
    Nov 12 at 1:07

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