1

I am mimicking the argument of Hatcher here from example $0.15$ of page no. 15. What the author says is as follows $:$

To verify the homotopy extension property, notice first that $I \times I$ retracts to $(I \times \{0\}) \cup (\partial I \times I),$ hence $B \times I \times I$ retracts to $(B \times I \times \{0\}) \cup (B \times \partial I \times I),$ and this retraction induces a retraction of $M_f \times I$ onto $(M_f \times \{0\}) \cup ((A \cup B) \times I).$

It's not quite clear to me that how does this induced retraction is obtained. I don't know why Hatcher explained important results reluctantly without any rigour. I am trying to think in terms of the following pushout diagrams. The mapping cylinder $M_f$ of $f$ can be given in terms of the following pushout diagram.

$$\require{AMScd} \begin{CD} B @>{i\ :\ b\ \mapsto\ (b,0)}>{}> B \times I \\ @V{f}VV @VV{\overline {f}}V\\ A @>{\overline i}>{\text {}}> M_f \end{CD}$$

The above pushout diagram gives rise to a pushout diagram of the following form $:$

$$\require{AMScd} \begin{CD} B \times I @>{i \times \text {id}_I\ :\ (b,t)\ \mapsto\ (b,0,t) }>{}> B \times I \times I \\ @V{f \times \text {id}_I}VV @VV{\overline f \times \text {id}_I}V\\ A \times I @>{\overline i \times \text {id}_I}>{\text {}}> M_f \times I\end{CD}$$

So in order to get a map $r : M_f \times I \longrightarrow (M_f \times \{0\}) \cup ((A \cup B) \times I)$ we need only to get hold of maps $j_1 : B \times I \times I \longrightarrow (M_f \times \{0\}) \cup ((A \cup B) \times I)$ and $j_2 : A \times I \longrightarrow (M_f \times \{0\}) \cup ((A \cup B) \times I)$ such that $j_1 \circ (i \times \text {id}_I) = j_2 \circ (f \times \text {id}_I).$

Would anybody please help me finding $j_1$ and $j_2\ $? If there is any easier way to approach the problem, I would be happy to learn that also.

Thank you very much.

Anacardium
  • 2,716
  • @Sumanta what do you mean by the map $i_0$ to have the homotopy extension property? As far as I know it turns out that HEP is satisfied by a pair $(X,A)$ such that whenever we have a map $f : X \longrightarrow Y$ and a homotopy $H : A \times I \longrightarrow Y$ such that $H(-,0) = f \big\rvert_{A}$ then there exists a homotopy $\widetilde H : X \times I \longrightarrow Y$ such that $\widetilde H \big\rvert_{A \times I} = H$ and $\widetilde H(-,0) = f.$ Am I right? Do you mean that the pair $(M_f, X \times {0})$ has HEP? – Anacardium May 19 '21 at 18:53
  • @Sumanta this problem is done from what you have answered for induced HEP for adjunction spaces. Because we already know that $(I,\partial I),$ has HEP since $I \times I$ retracts to $(I \times {0}) \cup (\partial I \times I)$ (we could have taken the radial projection from the point $\left (\frac {1} {2}, 2 \right )$ onto the three boundaries of the unit square (except the boundary on the top). – Anacardium May 21 '21 at 03:03
  • This shows that $(X \times I, X \times \partial I)$ has HEP. But now from what you have proved it follows that the pair of adjunction spaces $\left (Y \cup_f (X \times I), Y \cup_f (X \times \partial I) \right )$ has HEP. Now by definition $(X \times I) \cup_f Y = M_f$ and from the construction it is clear that $X \times {0}$ is identified with the points in $Y.$ So $X \times {0} \subseteq Y$ and $X \times {1}$ is the copy of $X$ sitting inside $X \cup_f Y.$ So this shows that $\left (M_f, X \cup Y \right )$ has HEP. Does it make sense what I have argued @Sumanta? – Anacardium May 21 '21 at 03:09
  • @Sumanta I may be missing something. I think what you have shown is that $(M_f, X)$ has HEP. Does it imply that $(M_f, X \cup Y)$ also has HEP? – Anacardium May 21 '21 at 06:37
  • Given a map $f : X \times I \longrightarrow Y$ and given a homotopy $\mathcal H : X \times I \longrightarrow Y$ of $f \big\rvert_{X}$ we can freeze the time interval at some time $s$ to get maps $\mathcal H(-,s) : X \longrightarrow Y$ and $f(-,s) : X \longrightarrow Y.$ Now the pair $(X,X)$ trivially has HEP. So this induces a map $h_s : X \times I \longrightarrow Y.$ Now we define $\widetilde H ((x,t),s) : = h_s (x,t).$ By gluing lemma it follows that $\widetilde H$ is continuous and thus we are through. Isn't it fine @Sumanta? – Anacardium May 21 '21 at 07:08
  • The above argument shows that the pair $(X \times I, X)$ has HEP. – Anacardium May 21 '21 at 07:14
  • Sorry, I misread your question. Here is the thing you are probably looking for. Let $f\colon B\to A$. There is a retraction $r=(r_1,r_2)\colon I\times I\to I\times 0\cup \partial I\times I\subseteq \Bbb R^2$ considering radial projection from $\left(\frac12,2\right)$.

    Now, we have a retarction $\text{Id}_B\times r\colon B\times I\times I\to B\times (I\times 0\cup \partial I\times I)=B\times I\times 0\cup B\times \partial I\times I$ given by $(\text{Id}_B\times r)(b,s,t)=\big(b,r(s,t)\big)=\big(b,r_1(s,t),r_2(s,t)\big)$.

     – Sumanta May 21 '21 at 10:20
  • Let $M_f=\frac{A\sqcup(B\times I)}{f(b)\sim(b,1)}$ with quotient map $q\colon A\sqcup(B\times I)\to M_f$, and define $A'=\big{[a]\in M_f: a\in A\big},$ $B'=\big{[b,0]\in M_f:b\in B\big}.$

    Consider the retraction $R\colon M_f\times I\to M_f\times 0\cup (A'\cup B')\times I$ given by $$R\big([b,s],t\big)=\bigg(\big[b,r_1(s,t)\big],\ r_2(s,t)\bigg),$$$$R\big([a],t\big)=\big([a],t\big).$$ Tell me if this is fine.

     – Sumanta May 21 '21 at 10:20
  • @Sumanta I have some doubt regarding continuity of the map $R.$ When a map is defined on an identification space we usually show it's continuity via universal property of quotient topology. Right? Here I am trying to think in that way. It is clear that $q \times \text {id}_I$ is continuous since $I$ is locally compact and Hausdorff. From here how do I induce $R\ $? – Anacardium May 21 '21 at 16:44
  • By this we have $M_f\times I\cong \frac{\displaystyle\big((B\times I)\sqcup A\big)\times I}{\displaystyle (b,1,t)\sim(f(b),t)}$. – Sumanta May 21 '21 at 17:20
  • Consider the quotient map $q'\colon((B\times I)\sqcup A)\times I\to M_f\times I$. Now, $R\circ q'=q'\circ \varphi$ where $\varphi\colon((B\times I)\sqcup A)\times I\to ((B\times I)\sqcup A)\times I$ as $$\varphi(b,s,t)= \big(b,r_1(s,t),r_2(s,t)\big)\text{ and }\varphi(a,t)=(a,t).$$ Notice that $((B\times I)\sqcup A)\times I\cong \big((B\times I)\times I\big)\sqcup (A\times I)$ and show $\varphi$ is continuous. From $R\circ q'=q'\circ \varphi$ we have $R$ is continuous as $q'$ is a quotient map. – Sumanta May 21 '21 at 17:20
  • Yeah I get it on my own. Here $\varphi$ is continuous by gluing lemma. – Anacardium May 21 '21 at 17:44

0 Answers0