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I guess I could use the property: The homotopy extension property for $(X,A)$ is equivalent to $X\times \{0\}\cup A\times I\ $ being a retract of $X\times I$.
Then there is a retraction:$$X\times I\rightarrow X\times \{0\}\cup A\times I$$
I need to show$$(X\cup CA)\times I\rightarrow (X\cup CA)\times \{0\}\cup CA\times I$$is a retraction too. But I don't know what to do next. Thank you for your time

Stefan Hamcke
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user12580
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2 Answers2

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Putting it more generally, if $(X,A)$ has the HEP, and $f:A \to B$ is a map, then $(B\cup_f X,B)$ has the HEP. The proof is analogous to that of Stefan, which is the case $B=CA$.

You need the method of defining homotopies on adjunction spaces, which again uses the local compactness of $I$.

Ronnie Brown
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Note that $(X,A)$ having the HEP implies that $X×\{0\}\cup A×I$ has the coherent topology with respect to the covering $(X×\{0\},A×I)$, that means a function from that space is continuous if its restrictions to $X×\{0\}$ and to $A×I$ are continuous on the respective spaces (so that space is homeomorphic to $M_i$, the mapping cylinder of the inclusion $i:A\to X$).

Consider the following commutative diagram $$ \begin{array} @ X\times I \sqcup CA×I & \longrightarrow & (X∪CA)×I \\ \downarrow r\sqcup 1_{CA} & & \downarrow r' \\ (X×\{0\}\cup A×I) \sqcup CA×I & \longrightarrow &(X∪CA)×\{0\}∪CA×I \end{array} $$ Since $I$ is locally compact, the upper map is a quotient map identifying $$ \qquad\qquad (a,t)\sim([a,0],t) \qquad\qquad (1) $$ for any $a\in A$ and $t\in I$. The map $r\sqcup 1_{CA×I}$ sends $$(x,t)\mapsto r(x,t),\quad ([a,s],t)\mapsto ([a,s],t) $$ where $r$ is the retraction. Finally, the lower map is defined on $X×\{0\}\cup A×I$ by sending $$ (x,0) \mapsto (x,0) \in (X\cup CA)×\{0\}, \quad (a,t) \mapsto ([a,0],t) \in CA×I $$ and it is continuous by the remark at the beginning of my post. Since the identifications (1) are respected by the lower left path, the map $r'$ is induced.

Stefan Hamcke
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  • The answer seems quite hard for me right now. I'll check it every week till I succeed to understand it.Though it makes me suffer… – user12580 Oct 28 '15 at 13:34
  • @Mr.Rock: I edited a bit. – Stefan Hamcke Oct 28 '15 at 15:23
  • Your answer is really nice. I have one small question here. How do you conclude that $r'$ is also a retraction? Has it anything to do with the commutativity of the diagram? Thanks. – Anacardium Jun 01 '21 at 06:11
  • Yeah I understand it now. This indeed follows from the commutativity of the diagram. Very nicely written answer. Thanks. But one small point I want to mention which is that the bottom map should send $(x,0)↦[x,0].$ Because $X∪CA$ is actually a mapping cone of the inclusion $A \hookrightarrow X$ and hence it's an identification space. The points of the identification space are usually classes of points of some former space on which identifications are made. Isn't it so? – Anacardium Jun 01 '21 at 07:00