# 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).

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!

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.

1. $$X$$ is perfectly normal and $$A\subseteq X$$ is closed.
2. $$X$$ is normal and $$A\subseteq X$$ is a closed $$G_\delta$$-set.
3. $$X$$ 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.

• Thank you so so much! This fills quite a hole in my current understanding of algebraic topology... Feb 17, 2020 at 16:59
• In your proof 3) => 4), isn‘t $t > u(x)$ but $h(t, x) \notin A$ possible if for example $t < 1$?
– HDB
Nov 12, 2021 at 1:07
• Is there a typo in the definition of $u$? $r$ being a retraction means that $r(0,x)=(0,x)$, so $pr_1(r(0,x))=0$. Jan 7 at 21:35
• @HDB There was a mistake in the original formulation. My apologies to anyone who did not notice it. I wrote this three years ago, and I guess I learned something in that time. Jan 8 at 5:14
• @PatrickNicodemus This has been fixed. I answered the question you linked by providing counterexamples here. Jan 8 at 5:14