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Let the pair $(X,A)$ have homotopy extension property. Let $B$ be a closed subset of $A$ and consider the map $\varphi : B \longrightarrow Y.$ Then can we say that the pair of adjunction spaces $\left (Y \cup_{\ \varphi} X, Y \cup_{\varphi} A \right )$ also have homotopy extension property? If so, how do I prove that? A small hint will be warmly appreciated at this stage.

Thanks in advance.

Anacardium
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  • At first, show that for a locally compact Hausdorff space $Z$ and a map $f\colon A\to Y$ we have $(X\cup_f Y)\times Z\cong (X\times Z)\cup_{f\times \text{Id}_Z} (Y\times Z)$. To prove this you need to use $X\cup_f Y$ as a pushout and $(X\sqcup Y)\times Z\xrightarrow{ \text{quotient}\times\text{Id}_Z} (X\cup_f Y)\times Z$ is also a quotient map.

    Now, for $Z=[0,1]$ from the previous lines we have $(X\cup_f Y)\times I$ is pushout of $f\times \text{Id}_Z\colon A\times I\to Y\times I$. Now, use the fact $(X,A)$ has HEP to conclude your result. Let me know if you need a full explanation.

     – Sumanta May 20 '21 at 09:31
  • @Sumanta you are applying Whitehead's theorem to conclude that $q \times \text {id}_Z$ is a quotient map as we are assuming that $Z$ is a locally compact Hausdorff space. Am I right? So what we can conclude at last is that the pair $\left ((X \cup_f Y) \times I, Y \times I \right )$ has HEP. – Anacardium May 20 '21 at 09:40
  • @Sumanta Yes that is exactly what is known as Whitehead's theorem. – Anacardium May 20 '21 at 09:44
  • But I am unable conclude the required result @Sumanta. – Anacardium May 20 '21 at 09:49

1 Answers1

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Lemma: For $A\subseteq X$ and $f\colon A\to Y$ we have $(X\cup_f Y)\times Z\cong (X\times Z)\cup_{f\times \text{Id}}(Y\times Z)$ for a locally compact Hausdorff space $Z$.

Proof: Consider the maps $i\colon A\hookrightarrow X$, $\kappa\colon X\to X\cup_f Y$, and $\ell\colon Y\to X\cup_f Y$. We need to show $(X\cup_f Y)\times Z$ is the pushout of the inner square below. That is given $G, F$ such that outer square is commutative we have to find $H$. enter image description here

But, for each $z\in Z$, we have the following diagram as $X\cup_f Y$ is a pushout. enter image description here

Now define $H(p,z):=h_z(p)$ for $p\in X\cup_f Y$ and $z\in Z$. And to show $H$ is indeed continuous consider the quotient map $q\colon X\sqcup Y\to X\cup_f Y$. Since $Z$ is LCH $q\times \text{Id}_Z$ is also a quotient map, so we are done by the following diagram.

enter image description here

Now, we want to show $\ell\colon Y\to X\cup_f Y$ has HEP for any $f\colon A\subseteq X\to Y$, provided $i\colon A\hookrightarrow X$ has HEP.

That is given $\alpha$ and $\mathcal H$ as in the diagram below such that outer square is commutative, we have to find $\widetilde {\mathcal H}$.

enter image description here

Note that $i\colon A\hookrightarrow X$ has HEP. So, we have the following:

enter image description here

Finally, $(X\cup_f Y)\times I$ is a pushout by the previous lemma. So, we have the diagram.

enter image description here

Check that this $\widetilde {\mathcal H}$ is what we need, i.e., show that it fits into the fourth diagram.

So, we have shown $(X\cup_f Y,Y)$ has HEP for any $f\colon A\subseteq X\to Y$, provided $A\hookrightarrow X$ has HEP. Now, to answer your question, note that the composition of two pushout squares is the pushout square. In other words, $ (Y\cup_\varphi A)\cup_\overline \varphi X\cong Y\cup_\varphi X $, i.e. $Y\cup_\varphi X$ is obtained as adujction space w.r.t. $\overline \varphi\colon A\to Y\cup_\varphi A$. So, $(X\cup_\varphi Y, A\cup_\varphi Y)$ has HEP.

enter image description here

Sumanta
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  • Very nice elaborate answer. Many many thanks. Can you suggest me some reference where we can find arguments using pushouts? I have found it nowhere in Hatcher's book. – Anacardium May 20 '21 at 15:59
  • This book is useful for the diagrammatic interpretation of basic classical homotopy theory, but it does not contain any homology or cohomology theory. Also, proving "composition of two pushout squares is again a pushout square" is very straightforward. – Sumanta May 20 '21 at 16:32
  • Just a small question. The second diagram what you have drawn to show that the pair $(X \cup_f Y, Y)$ has HEP, there we need to verify whether $(\mathcal H \circ (f \times \text {id})) (a,0) = (\alpha \circ \kappa) (a)$ or in other words $\mathcal H (f(a),0) = (\alpha \circ \kappa) (a).$ Right? Although it is very easy to show. Because from the first diagram it follows that $$\mathcal H(f(a),0) = (\alpha \circ \ell) (f(a) = \alpha (\ell (f(a)) = \alpha (\kappa (a)) = (\alpha \circ \kappa) (a).$$ The penultimate equality follows from the commutativity of the pushout diagram $X \cup_f Y.$ – Anacardium May 20 '21 at 16:38
  • Is there any free soft copy of this book available online? – Anacardium May 20 '21 at 16:41
  • Right. You need to check every diagram is commutative, the small squares, triangles, etc. Keep in mind in $X\cup_f Y$ we have $[a]=\big[f(a)\big]$ for all $a\in A$. I don't know about the freely available soft copy of this book. – Sumanta May 20 '21 at 16:48