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I'm looking for a pair $(X,x)$ consisting of a topological space $X$ with a distinguised point $x \in X$ such that canon inclusion $i_x: x \to X$ is not a cofibration, ie not satisfied homotopy extension property.

There are known several examples for pairs of spaces $(X,A)$ (see eg here or here) violating HEP,
but I would like to see one where $A$ is really just a single point.

Are there also a "heuristic/intuitive" reasons what potentially obstucts/prevents a homotopy $H_A:\{x\} \times I \to T$ with prescribed map $f: X \to T$ to admit such an extension $H_X: X \times I \to T$ with $H_X \vert _{A \times I}=H_A$ and $H_0=f$? ($T$ some test space)

Which picture one should keep in mind seeking for a counterexample? Clearly $(X,x)$ isn't allowed to be a CW-pair, so it becomes rather "weird".

Inspired by this I would guess (but was till now without success in proving it) that $X = \{0\}\cup\left\{\frac{1}{n} \ : \ n \in\mathbb{N}\right\}$ with $A= x=\{0\}$ (with subspace topology inherited from $\Bbb R$) could have desired properties.
Then a homotopy $H_A:\{x\} \times I \to T$ corresponds simply to a path $p_x: I \to T$. So the question is to find a weird enough "test space" $T$ and a map $f: X \to T$ to it "entangling" bad enough such that potential homotopy $H_X: X \times I \to T$ cannot garantee to restrict to $H_A$ and $f$ simultaneosly.
Is it possible to produce with such $X$ the desired counter example?

user267839
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    Have you considered taking the first example that you posted (where $X=[0,1]$ and $A$ is the terms of the harmonic series union their limit $0$) and passing to the quotient pair $(X/A, )$ (where the point $$ is the image of $A$ under the quotient map)? I don't know for a fact that this will work, but it would be my first guess given that example. – Tabes Bridges Dec 12 '24 at 19:32
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    You can at least use your $X$ to build a counterexample. See Example 8.31 in these notes. – Ben Steffan Dec 12 '24 at 19:40
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    There is always universal `test space' to test the HEP. It's the first place to look, and in this case probably the most accessible. Namely, an inclusion $A\subseteq X$ has the HEP if and only if $X\times0\cup A\times I$ is a retract of $X\times I$. It's easy to show that this fails for the pair $(X,{0})$ which you consider. – Tyrone Dec 12 '24 at 19:44
  • @Tyrone: presumably I see. I terms of HEP one considers $T:=X\times0\cup A\times I$ with $f: X \to T, H_A: A \times I \to T$ canonical inclusions. Then the hypothetical extension $H_X: X \times I \to T$ would be discontinous with resp sequence $(1/n,1)_n \to (0,1)$. That's the key problem, right? – user267839 Dec 12 '24 at 19:55
  • You should write an official answer to your own question. – Paul Frost Dec 24 '24 at 09:40
  • @PaulFrost: Ok, but intended not to attribute Tyrone's merit to oneself – user267839 Dec 24 '24 at 12:01
  • @user267839 In your answer you correctly attributed the idea to Tyrone. Your merit is that you cleared the question from the unanswered queue. – Paul Frost Dec 26 '24 at 00:32

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following Paul Frost's advice let me just repeat Tyrone's idea (...at least how I understood him) to complete this issue:

We consider as "test space" $T:=X\times0\cup A\times I$ with $f: X \to T, H_A: A \times I \to T$ canonical embeddings. The problem is that any hypothetical extension $H_X: X \times I \to T$ cannot be continuous in $(0,1)$: Consider the sequence $(1/n,1)_n \to (0,1)$ and another sequence $(a_n, 1)_n \to (0,1)$ where $a_n \to 0$ are disjoint from the $1/n$'s, eg all irrational. Then by construction, both converge to $(0,1)$, but $(H_X(a_n, 1))_n$ not converge to $(0,1)$ in contrast to $ ((H_X(1/n,1))_n=(1/n,1)_n$ as it should for continuous map

user267839
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Here is a simpler example. Take $X= \{a,b\}$ with open sets $\emptyset$, $\{b\}$ and $X$. For some inclusion of a closed subset $A \hookrightarrow X$ to be a cofibration, there must be a continuous map $\phi\colon X \to [0,1]$ such that $A = \phi^{-1}(\{0\})$ (this is a necessary condition). For $\{a\}$ to be $\phi^{-1}(\{0\})$ you have no choice for $\phi$: $\phi(a) = 0$ and $\phi(b) = t \not= 0$. But then $\phi$ is not continuous because $\phi^{-1}([0,\frac{t}{2}[) = \{a\}$ which is not an open set of $X$.

jpaul
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