3

Assume $X$ is a nice enough metric space, if $(X,A)$ is space , $A$ closed in $X$, such that for some $U$ open in $X$, with $A\subseteq U$, $(U,A)$ is an NDR pair, does it follow that $(X,A)$ is an NDR pair?

Here, $(X,A)$ NDR means that there exists a continuous map $\phi:X\rightarrow [0,1]$ such that $A=\phi^{-1}(0)$ and a homotopy $H:X\times I\rightarrow X$ rel $A$ such that $H$ starts at the identity and for $t>\phi(x)$, $H(x,t)\in A$.

Now if $(U,A)$ is an NDR and are above, it is not clear to me how one extends $H:U\times I\rightarrow U$ and $\phi:U\rightarrow I$ to maps $H':X\times I\rightarrow X$ and $\phi':X\times I\rightarrow X$.

If say $(X\times 0) \cup (A\times I)\subseteq U\times I$ then it is clear how to do this, but cannot assume this in general.

dsh
  • 1,356
monoidaltransform
  • 554
  • Suppose $X$ is a circle and $A$ consists of just one point. Small intervals $U$ around $A$ can be continuously deformed into $A$, but the whole circle cannot. – Andreas Blass Sep 18 '24 at 18:05
  • 2
    @AndreasBlass This question is about neighborhood retracts/closed cofibrations by Strom's characterization. $(0, S^1)$ is a cofibration. – dsh Sep 18 '24 at 18:49

1 Answers1

3

The following two theorems are classical characterisations of the absolute HEP.

Theorem (Strøm) A closed subspace inclusion $j:A\hookrightarrow X$ is a cofibration if and only if there exists a map $\varphi:X\rightarrow I$, and a homotopy $H:X\times I\rightarrow X$, satisfying

  1. $A=\varphi^{-1}(0)$,
  2. $H_0=id_X$
  3. $H_t|_A=id_A$ for all $t\in I$
  4. $H(x,t)\in A$ whenever $t>\varphi(x)$. $\quad\blacksquare$

Theorem An inclusion $A\subseteq X$ is a cofibration if and only if $A\times I\cup X\times 0$ is a retract of $X\times I$. $\quad\blacksquare$

Putting these together allows for a new characterisation (which is also fairly well-known).

Theorem Let $X$ be a space and $A\subseteq X$ a closed subset. Then the inclusion $j:A\subseteq X$ is a cofibration if and only if $A$ has a neighbourhood $U\subseteq X$ for which there is a map $\varphi:X\rightarrow I$ and a homotopy $H:U\times I\rightarrow X$ such that

  1. $A=\varphi^{-1}(0)\subseteq\varphi^{-1}[0,1)\subseteq U$.
  2. $H_0(x)=x$ for all $x\in U$
  3. $H_t(a)=a$ for all $a\in A$, $t\in I$
  4. $H_1(x)\in A$ for all $x\in U$.

Proof For necessity assume $j:A\hookrightarrow X$ is a cofibration and fix a pair $(\varphi,H)$ as in Strøm's Theorem. Putting $U=\varphi^{-1}[0,1)$, the triplet $(U,\varphi,H)$ satisfies the requirements of the statement.

For sufficiency we define a retraction $r:X\times I\rightarrow X\times0\cup A\times I$ by $$ r(x,t)=\begin{cases} (x,t)&x\in\varphi^{-1}(0)\\ (H(x,1),t-2\varphi(x))&x\in\varphi^{-1}(0,1/2]\;\text{and}\;2\varphi(x)\leq t\leq1\\ (H(x,t/2\varphi(x)),0)&x\in\varphi^{-1}(0,1/2]\;\text{and}\;t\leq2\varphi(x)\\ (H(x,2(1-\varphi(x))t),0)&x\in \varphi^{-1}[1/2,1)\\ (x,0)&x\in \varphi^{-1}(1). \end{cases} $$ $\blacksquare$

The following is the desired application. Recall that a space is perfectly normal if it is normal and each of its closed subsets if $G_\delta$. Equivalently, $X$ is perfectly normal if for any pair of disjoint closed subsets $A,B\subseteq X$ there is a continuous function $\varphi:X\rightarrow I$ such that $A=\varphi^{-1}(0)$ and $B=\varphi^{-1}(1)$. Clearly every metric space is perfectly normal. Moreover, one can show that perfect normality is preserved by adjunction spaces and sequential colimits, so every CW complex is perfectly normal.

Corrollary Let $X$ be a perfectly normal space and $A\subseteq X$ a closed subset. Assume that $U\subseteq X$ is a neighbourhood of $A$ and the inclusion $A\subseteq U$ is a cofibration. Then $A\subseteq X$ is a cofibration.

Proof Since $A\subseteq U$ is a cofibration, by the previous theorem there is a neighbourhood $V\subseteq U$ of $A$ and a homotopy $H:V\times I\rightarrow U$ satisfying the listed properties.

On the other hand, since $X$ is perfectly normal, $A$ is closed in $X$, and $V$ is a neighbourhood of $A$ in $X$, there is a map $\varphi:X\rightarrow I$ such that $A=\varphi^{-1}(0)\subseteq\varphi^{-1}[0,1)\subseteq V$. The triplet $(V,\varphi,H)$ satisfies the requirements of the theorem, so $A\subseteq X$ is a cofibration. $\blacksquare$

Tyrone
  • 17,539
  • hello could you elaborate on how one constructs maps like $r$? They seem like magic to me. – monoidaltransform Sep 19 '24 at 18:14
  • can I have a reference for the second theorem ? Im having trouble understanding intuition behind the construction of r – monoidaltransform Sep 20 '24 at 09:12
  • Hi @monoidaltransform. I was shown the retraction some time ago, and I'm probably pretty bad at explaining intuition. I'd start by sketching $X\times I$ as a 2-dimensional region and juxtaposing over it the graph of $\varphi$ and the cylinders $A\times I$ and $U\times I$. The area above the graph of $\varphi$ looks like fattening up of $A\times I$ in $U\times I$. The region $\varphi^{-1}[0,1/2)\times I$ is broken up into two pieces according to whether the points of $X$ sit above or below the graph of $\varphi$. – Tyrone Sep 20 '24 at 12:29
  • The second part of $r$ pushes the points above the graph into $A\times I$, and the third part of $r$ collapses the points under the graph to $X\times 0$. The fourth part of $r$ blends the `non linear' deformation provided by the continuous $H$, with the collapse to $X\times 0$ that must occur outside of $U$ for $r$ to be well-defined. – Tyrone Sep 20 '24 at 12:30
  • Thank you, one more question, why is $r$ continuous? – monoidaltransform Sep 20 '24 at 15:48
  • Continuity is clear away from the points $A\times I$ (maybe you are having trouble with other points also?). Thus take $(x_0,t_0)\in A\times I$. Note that $x_0=H(x_0,1)$. Assume first that $t_0>0$. Then there is $0<s_0<t_0$ and by continuity of $\varphi$ a nhd $V$ of $x$ such that $\varphi(V)\subseteq[0,s_0/2)$. Then $r|_{V\times(s_0,1]}(x,t)=(H(x,1),t-2\varphi(x))$. Thus $r$ is continuous in a nhd of $(x_0,t_0)$, hence continuous at $(x_0,t_0)$. – Tyrone Sep 21 '24 at 13:31
  • Now assume $t_0=0$. Since $r(x_0,t_0)=(x_0,t_0)$ it is enough to show that for any nhd $W$ of $x_0$ and any $\epsilon>0$ there is a nhd $V$ of $x_0$ such that $r(V\times[0,\epsilon))\subseteq W\times[0,\epsilon)$. For each $s\in I$ it holds that $H(x_0,s)=x_0$. Thus let $V_s$ be a nhd of $x_0$ and $s\in J_s\subseteq I$ an open subinterval such that $H(V_s,J_s)\se W$. By compactness, $I=\bigcup_{i=1}^n J_{s_i}$ for some $s_1,\dots,s_n\in I$. Put $V=\bigcap^n_{i=1}V_{s_i}$. Then $H(V\times I)\subseteq W$, and hence $r(V\times[0,\epsilon)\subseteq W\times[0,\epsilon)$. – Tyrone Sep 21 '24 at 13:32