# 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 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 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!

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.

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