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!


1 Answer 1


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.

  • $\begingroup$ Thank you so so much! This fills quite a hole in my current understanding of algebraic topology... $\endgroup$ Commented Feb 17, 2020 at 16:59
  • 1
    $\begingroup$ In your proof 3) => 4), isn‘t $t > u(x)$ but $h(t, x) \notin A$ possible if for example $t < 1$? $\endgroup$
    – HDB
    Commented Nov 12, 2021 at 1:07
  • $\begingroup$ Is there a typo in the definition of $u$? $r$ being a retraction means that $r(0,x)=(0,x)$, so $pr_1(r(0,x))=0$. $\endgroup$ Commented Jan 7, 2023 at 21:35
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    $\begingroup$ @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. $\endgroup$
    – Tyrone
    Commented Jan 8, 2023 at 5:14
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    $\begingroup$ @PatrickNicodemus This has been fixed. I answered the question you linked by providing counterexamples here. $\endgroup$
    – Tyrone
    Commented Jan 8, 2023 at 5:14

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