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A pair of topological spaces $(X, A)$ is a cofibred pair if $A$ is closed in $X$ and the homotopy extension property holds. Moreover, $(X, A)$ is a good pair if $A$ is closed in $X$ and there exists an open neighbourhood $U$ of $A$ such that $A$ is a deformation retract of $U$. In Good pair vs. cofibration the reader can find an example of a good pair that is not cofibred and vice-versa.

If $X$ is compactly generated and $(X, A)$ is cofibred, then necessarily $A$ is a $G_{\delta}$ subset of $X$, since $(X, A)$ is a NDR-pair (see Steenrod, "A convenient category of topological spaces", sections 6 and 7). In the good pair $(X, A)$ shown in Good pair vs. cofibration, that is not cofibred, the subset $A$ is not $G_{\delta}$. I would like to see an example of a good pair $(X, A)$ such that $A$ is $G_{\delta}$ and $(X, A)$ is not cofibred.

Fabio
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  • Just a remark, but by this other answer by Tyrone, a counter-example requires $X$ to not be normal (in particular, it cannot be metrizable or compact Hausdorff), if not worse. – Thorgott Nov 24 '23 at 21:09
  • I have a doubt about the answer you quoted. The condition (*) there is claimed to hold if X is normal and A is $G_{\delta}$. Nevertheless, we can equivalently state that condition in the form $A = v^{-1}(0)$ and $X \setminus N = v^{-1}(1)$, hence $X \setminus N$ has to be $G_{\delta}$ too. If $X$ is perfectly normal, then this is automatic, but if $X$ is "only" normal, then it is not. – Fabio Nov 26 '23 at 10:41
  • Hence, the only implication that is clear to me is the following one: if $X$ is perfectly normal, then $(X, A)$ good pair implies $(X, A)$ cofibred (in this case $A$ is automatically $G_{\delta}$). The converse does not hold, since the counter-example in https://math.stackexchange.com/questions/854281/good-pair-vs-cofibration is a metric space. Now my questions are: (1) If $X$ is normal and $A$ is $G_{\delta}$, then is it true that $(X, A)$ good pair implies $(X, A)$ cofibred? (2) Is there a "reasonable" category (e.g. CW-complexes) in which good pair is equivalent to cofibred? – Fabio Nov 26 '23 at 12:41
  • This might need a small extra argument. Condition (2) is actually preserved when replacing $N$ with a smaller neighborhood $N^{\prime}$ of $A$ in $X$. So it suffices to find a $v$ s.t. $v^{-1}(0)=A$ and $v^{-1}([0,1))\subseteq N$. However, if you have $N$ and $v$ s.t. $v^{-1}(0)=A$, Urysohn's lemma also lets you pick a $\varphi\colon X\rightarrow I$ s.t. $\varphi\vert_A=0$ and $\varphi\vert_{X\setminus N}=1$. Then, if you set $\psi=\max{\varphi,\nu}$, you have $\psi^{-1}(0)=A$ and $\psi^{-1}([0,1))\subseteq N$. So, if $X$ is normal and $A$ is closed $G_{\delta}$, we have $(2)$ iff cofibered. – Thorgott Nov 26 '23 at 13:23
  • Yes, it's ok. I got confused because, when $(X, A)$ is a good pair, the homotopy is of the form $F \colon N \times I \to N$, but in condition (2) it is only required that $F \colon N \times I \to X$, hence it trivially keeps on holding when reducing $N$. – Fabio Nov 26 '23 at 13:31

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