# Importance of well-pointedness (in particular for the pointed mapping cylinder construction)

In a recent question, I asked wether well-pointedness was indeed necessary to make the canonical inclusion $$X\hookrightarrow M_f^{\bullet}$$ a pointed cofibration. Here $X,Y$ are pointed topological spaces (compactly generated weak Hausdorff spaces to be precise), $f:X\to Y$ is a pointed map and $M_f^{\bullet}$ its pointed mapping cylinder. Unless I am deeply mistaken, the canonical inclusion is always a pointed cofibration (and the proof is virtually identical to the unpointed case), however, my main source on this material, Jeff Strom's Modern Classical Homotopy Theory, tells me that one should make the additional assumption of $Y$ being well-pointed, that is, the inclusion of the base point $\ast\hookrightarrow Y$ is an unpointed cofibration.

I could imagine well-pointedness of $Y$ being useful if we wanted to show $X\hookrightarrow M_f^{\bullet}$ to be an unpointed cofibration, although I couldn't show it actually is one. I've checked J.P. May's account on the subject in his A concise Course in Algebraic Topology, and well-pointedness doens't seem to play nearly as big a role as in Strom's book, at least May puts less emphasis on this condition.

What's the deal with well-pointedness? Am I wrong about $X\hookrightarrow M_f^{\bullet}$ necessarily being a pointed cofibration? Why would I need well-pointed spaces, when in the pointed context I am only interested in extending pointed homotopies? Lastly, is it true that when $Y$ is well pointed, the canonical inclusion $X\hookrightarrow M_f^{\bullet}$ is an unpointed cofibration?

I am aware of a theorem that states that if $(\ast\in )A\rightarrow X$ is a pointed map and $A$ is well-pointed, then $A\rightarrow X$ is a pointed cofibration iff it is an unpointed cofibration. This seems like one important benefit of well-pointedness, but it doen't relate to my problem at hand.

EDIT @JustinYoung Thank you for your patience with this. I am not convinced by your counter example, here's why: the pointed mapping cylinder of the identity of $X$ should be homeomorphic to the countable union of closed segments with extremities $(1/n,0,0)$ and $(1/n,1/n,0)$ plus the origin as a subspace of $\Bbb R^3$ (and also to $X\rtimes I$). Thus, $M^{\bullet}_{id}\rtimes I$ should be homeomorphic to the countable union of full squares with corners $(1/n,0,0),(1/n,1/n,0),(1/n,1/n,1/n),(1/n,0,1/n)$ plus the origin as a subspace of $\Bbb R^3$ (and also to $(X\rtimes I)\rtimes I$). In this picture, $X\rtimes I\cup_X M_{id}^{\bullet}\rtimes\{0\}$ is all that has either $x=y$ or $z=0$. If you compress all these squares simultaneously onto their bottom right corner and sides, you get a strong deformation retraction of $M^{\bullet}_{id}\rtimes I$ onto $X\rtimes I\cup_X M_{id}^{\bullet}\rtimes\{0\}$. If you want I can can add my proof of the fact that the canonical inclusion $X\hookrightarrow M_f^{\bullet}$ is always a pointed cofibration, regardless of wether $Y$ is well pointed or not, and you can tell me if you agree or disagree. I can't to it tonight, but I might get to it tomorow.

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As for the first question, let $X$ be the basepoint of $Y$ and let $f$ be the inclusion $X = * \to Y$. Then, the mapping cylinder $M_f = Y$ and the inclusion $X \to M_f$ is just the inclusion $X = * \to Y$. This is clearly a cofibration if and only if the base point is nondegenerate. –  Justin Young Jul 26 '12 at 22:23
@JustinYoung Thanks for the comment. I'm not sure which question you answer though, what question is this an answer to? –  Olivier Bégassat Jul 26 '12 at 22:58
The question is whether $X \to M_f$ is a pointed fibration, when $Y$ is not well pointed. –  Justin Young Jul 27 '12 at 0:35
@JustinYoung I'm pretty sure it is a pointed cofibration regardless of wether $Y$ is well pointed or not: the definitions of Strom and May asks for the extendability of homotopies in the pointed category: during the whole process, the base point is always sent to the base point of the target space, so we don't need well pointedness of $Y$ as the base point is not allowed free roam. –  Olivier Bégassat Jul 27 '12 at 1:15
Consider the example $X = \{0\}\cup \{1/n \mid n>0\}$, with basepoint $0$, and form the mapping cylinder of $id: X \to X$. Then, you can show that $X \to M_{id}$ is not a pointed cofibration by showing that $M_{id} \rtimes I$ does not retract onto $X\rtimes I \cup_X M_{id} \rtimes \{0\}$. –  Justin Young Jul 31 '12 at 2:03