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.
- $X$ is perfectly normal and $A\subseteq X$ is closed.
- $X$ is normal and $A\subseteq X$ is a closed $G_\delta$-set.
- $X$ 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.
Suppose that $i:A\subseteq X$ is a subspace inclusion which is both a cofibration and a homotopy equivalence. We place no other assumptions on $X$ or $A$. The assumptions grant a map $r':X\rightarrow A$ such that $r'i\simeq id_A$ and $ir'\simeq id_X$. Using the homotopy extension property we obtain a map $r:X\rightarrow A$ such that $r\simeq r'$ and $ri=id_A$. Necessarily $ir\simeq id_X$.
$A$ is a deformation retraction of $X$.
Now fix a homotopy $F:id_X\simeq ir$. Define
$$G:X\times\{0,1\}\times I\cup A\times I\times I\rightarrow X$$
by putting
$$G(x,0,t)=x,\qquad G(a,s,t)=F(a,(1-t)s),\qquad G(x,1,t)=F(r(x),1-t).$$
This is well-defined and continuous. We need the following.
Lemma: Let $X$ be a space and $A\subseteq X$ a cofibration. Then $X\times\{0,1\}\cup A\times I\subseteq X\times I$ is a cofibration. $\square$
The proof makes use of the fact that there is a self-homeomorphism of $I^2$ which maps $I\times\{0,1\}\cup\{0\}\times I$ onto $\{0\}\times I$. I'll leave details to your textbook.
Above we have defined $G$. Notice that its restriction to $(X\times\{0,1\}\cup A\times I)\times\{0\}$ is just the map $F$. Because of the lemma there exists a homotopy $\widetilde G:X\times I\times I\rightarrow X$ with $\widetilde G(x,s,0)=F(x,s)$ and $H|_{X\times\{0,1\}\times I\cup A\times I\times I}=G$. Put
$$H_t(x)=\widetilde G(x,t,1).$$
This yields $H_0=id_X$, $H_1=ir$ and $H_t|_A=i$. Thus we have our conclusion.
A subspace inclusion $A\subset X$ which is both a cofibration and a homotopy equivalence is a strong deformation retract.
Of course if $X$ is contractible, then the inclusion $x_0\hookrightarrow X$ of any point is a homotopy equivalence. If this inclusion is a cofibration, then the outcome above applies. The Hausdorff and normality assumptions are not necessary.
Best Answer
Call a pair $(X,A)$ cofibred if $A$ is a closed subspace of $X$ and the inclusion $A\subseteq X$ has the homotopy extension property. I will produce examples of $(i)$ a good pair which is not cofibred, and $(ii)$ a cofibred pair which is not good. Neither example is my own, but they are not as well known as they should be.
A good pair which is not cofibred. Let $\Omega$ be an uncountable cardinal and consider the Tychonoff cube $I^\Omega$. The inclusion $0\hookrightarrow I^\Omega$ of the origin is evidently a strong deformation retract. Therefore the pair $(I^\Omega,\{0\})$ is good. However the pair is not cofibred. For if it were the point $0$ would be the zero-set of a continuous function $\varphi:I^\Omega\rightarrow I$, and this point is not even a G$_\delta$-set.
Discussion: The space $I^\Omega$ is a locally contractible compact $T_2$ space (which is separable when $\Omega\leq 2^{\aleph_0}$). As such it serves as a counterexample both in the category of all spaces and in the category of compactly generated spaces. The fact that the deformation retract of $I^\Omega$ onto $0$ can be chosen to fix $0$ at all times means that there is a homotopy equivalence of pairs $(I^\Omega,0)\simeq(0,0)$. Stated another way this reads; being well-pointed is not an invariant of pointed-homotopy type.
A cofibred pair which is not good. Sergey Melikhov describes here Borsuk's example of a 2-dimensional, compact, contractible metric space $M$ possessing the singularity of Mazurkiewicz. The original example is found on page 152 of Borsuk's book The Theory of Retracts. Our counterexample is its subspace, and here are the salient properties of $M$ which we will need to find it.
$1)$ $M$ is an absolute retract for metric spaces. Consequently, if $U\subset M$ is any open subset and $x\in U$ any point, then the inclusion $x\hookrightarrow U$ is a cofibration.
$2)$ $M$ is locally contractible in the sense that for any open $U\subseteq M$ and any $x\in U$, there is an open $V\subseteq U$ with $x\in V$ for which the inclusion $V\hookrightarrow U$ is null-homotopic. Still, $M$ is not a countable union of its contractible proper subsets. Since $M$ is a compact metric space it is separable and first-countable, and consequently $M$ has points which do not possess a base of contractible neighbourhoods.
We turn to our construction. Fix any point $x_0\in M$ which does not have a base of contractible neighbourhoods and let $\{U_n\}_\mathbb{N}$ be a neighbourhood base at $x_0$. We can assume that the $U_n$'s form a decreasing sequence of noncontractible open sets.
Claim: There is $n\in\mathbb{N}$ such that $(U_n,x_0)$ is not a good pair.
This is verified by exhaustion. If $(U_1,x_0)$ is not a good pair, then we are done. Thus suppose that $(U_1,x_0)$ is a good pair. This means that $x_0$ has a contractible neighbouhood $V_1\subseteq U_1$. Then using the fact that the $U_n$'s form a neighbourhood base at $x_0$ we find $n\in\mathbb{N}$ such that $U_n\subseteq V_1$. Now repeat the argument with $(U_n,x_0)$. If this pair is good we stop, and if not we have $U_m\subseteq V_n\subseteq U_n$, where $m>n$ and $V_n$ is a contractible neighbourhood of $x$.
Now, this process must terminate to yield a non-good pair. For if it did not we could re-enumerate the $V_n$'s and $U_n$'s to obtain a sequence $\{V_n\}_\mathbb{N}$ of contractible open neighbourhoods of $x_0$ such that $V_n\subseteq U_n$ for each $n\in\mathbb{N}$. Because $\{U_n\}_\mathbb{N}$ is a base at $x_0$, so would be $\{V_n\}_\mathbb{N}$, and this contradicts the fact that $x_0$ does not have a base of contractible neighbourhoods. $\square$
The end result is the presence of a noncontractible open $X\subseteq M$ which contains $x_0$, and is such that $(X,x_0)$ is not good. But $X$, being an open subset of an absolute retract, is itself an absolute neighbourhood retract. As was discussed above, a consequence of this is that the inclusion $x_0\hookrightarrow X$ is a cofibration. Thus we have obtained a cofibred pair $(X,x_0)$ which is not good.
Discussion: The space $X$ is a locally contractible (in the above sense) separable 2-dimensional metric space. In any case $(X,x_0)$ is a counterexample in the categories of all spaces, compactly generated spaces, and $\Delta$-generated spaces.
Note that $X$ has the pointed homotopy type of a countable, finite-dimensional CW complex. Thus not even in the category of well-pointed spaces is being a good pair an invariant of pointed homotopy type.