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Suppose that there exist continuous maps $f:Y\longrightarrow Z$ and $g:Z\longrightarrow Y$ so that $g\circ f\simeq 1_Y$. Also, let $h :X\longrightarrow Y$ be a continuous map which induces an isomorphism of fundamental groups. We know that $M_{h}\simeq Y$ and $M_{f\circ h}\simeq Z$, where $M_{h}=\frac{X\times I \cup Y}{(x,1)\sim h (x)}$ is the mapping cylinder of $h$.

Is $\pi_2 (M_{h},X)$ a direct summand of $\pi_2 (M_{f\circ h},X)$?

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  • $\begingroup$ Have your tried to construct the map $G$ from the homotopy from $g\circ f$ to $\operatorname{id}_Y$, restricted to $X\times I$, and $g$ on $Z$? $\endgroup$
    – Sebastian Goette
    Sep 18, 2017 at 6:48
  • $\begingroup$ Could you explain a bit more? $\endgroup$
    – MHenry
    Sep 18, 2017 at 10:02

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I think you can do the following : factor your map $F\colon M_h\to M_{f\circ h}$ as the inclusion $i\colon M_h\hookrightarrow M_h\cup_Y M_f$ followed by a map $q\colon M_h\cup_Y M_f\to M_{f\circ h}$. Then, construct a retract $r\colon M_h\cup_Y M_f\to M_h$. It is then enough to prove that the map $M_h\cup_Y M_f\to M_{f\circ h}$ is a weak-equivalence.

Let us first construct the map $q\colon M_h\cup_Y M_f\to M_{f\circ h}$ and show that it is a weak-equivalence. Consider the pushout diagram defining $M_h\cup_Y M_f$, where $i_0(y)=(y,0)$.

$$\require{AMScd}\begin{CD} Y@>i_0>>M_f\\ @VjVV@VV\widetilde{j}V\\ M_{h}@>i>>M_h\cup_Y M_f\\ \end{CD}$$

To define $q$, it is enough to define two maps $q_f\colon M_f\to M_{f\circ h}$ and $q_h\colon M_h\to M_{f\circ h}$ such that $q_f\circ i_0= q_h\circ j$. Set $q_f$ as the composition $M_f\to Z\hookrightarrow M_{f\circ h}$ and define $q_h$ via $q_h(x,t)=(x,t), q_h(y)=f(y)$. This defines the desired map $q\colon M_h\cup_ Y M_f\to M_{f\circ h}$.

Now, $\widetilde{j}$ and $q_f$ are homotopy equivalences, and so $q$ must induce isomorphisms on all homotopy groups. Since $q_{|X}\colon X\hookrightarrow M_{f\circ h}$ is equal to the inclusion, we have that $q$ induces isomorphisms on all relative homotopy groups.

Now, we define $r\colon M_h\cup_Y M_f\to M_{h}$. As earlier, it is enough to define $r_f\colon M_f\to M_h$ and $r_h\colon M_h\to M_h$, such that $r_f\circ i_0=r_h\circ j$. Take $r_h=Id_{M_h}$ and define $r_f$ via : $r_f(y,t)= H(y,t)$, and $r_f(z)=g(z)$. This defines $r$ such that $r\circ i= Id_{M_h}$.

Putting everything together, we have that the composite $M_h\xrightarrow{i} M_h\cup_Y M_f\xrightarrow{r} M_h$ is equal to the identity, and $\pi_n(M_h\cup_Y M_f,X)\simeq \pi_n(M_{f\circ h},X)$. This should imply the result you want.

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  • $\begingroup$ Excuse me. By the answer, we have $q\circ i_f =q_h$ and $q\circ \tilde{j}=q_f$, But why $\tilde{j}$ and $i_f$ are homotopy equivalence? Why does $q$ induce isomorphisms on all homotopy groups? $\endgroup$
    – MHenry
    Jun 22, 2018 at 4:57
  • $\begingroup$ I think the map $\tilde{j}$ is homotopy equivalence not the map $i$. I don't understand your mean about the map $i_f$ in the context. $\endgroup$
    – MHenry
    Jun 22, 2018 at 12:00
  • $\begingroup$ I am sorry, I changed my mind on notations in the middle of the process of writing this answer, and forgot one occurence of $i_f$. $i_f$ is supposed to be $q_f$. $\endgroup$
    – S. Douteau
    Jun 22, 2018 at 21:13
  • $\begingroup$ We have $q_f=q\circ\widetilde{j}$. $q_f$ is an homotopy equivalence since $M_f$ and $M_{f\circ h}$ are both homotopy equivalent to $Z$, and $\widetilde{j}$ is also an homotopy equivalence. Then, by the two out of three property (or equivalently, passing to homotopy groups) we see that $q$ must be a weak equivalence. $\endgroup$
    – S. Douteau
    Jun 22, 2018 at 21:17
  • $\begingroup$ Thank you so much for your complete answer. I learned lots of things from your proof. Thanks. $\endgroup$
    – MHenry
    Jun 25, 2018 at 12:39
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This is a comment with a commutative diagram rather than a full answer.

Let $H\colon Y\times I\to Y$ be a homotopy between $\operatorname{id}_Y$ and $g\circ f$. I though defining $G$ using the diagram $$\require{AMScd}\begin{CD} X\times I@>H\circ(h\times\operatorname{id}_I)>>Y\times I\\ @VVV@VVV\\ M_{f\circ h}@>>>Y\times I@>r>\sim>M_h\\ @AAA@AAA\\ Z@>g>>Y\rlap{\;,} \end{CD}$$ where $r$ is a strong deformation retract from $Y\times I$ to $M_h$. This definitely works if $h$ is a cofibration, and it seems that one can use $H$ to construct the homotopy between $G\circ F$ and $\operatorname{id}_{M_h}$.

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  • $\begingroup$ Thank you for your help. I think your mean is that $H\circ (h\times id_I )$ is a map from $X\times I$ to $Y$ in the first row of the diagram. Excuse me, could you tell me what is the map $M_{f\circ h}\longrightarrow Y$ in the second row in the diagram? $\endgroup$
    – MHenry
    Sep 19, 2017 at 14:52
  • $\begingroup$ Actually, I also want to keep the variable $t\in I$, so the upper map actually is $(x,t)\mapsto(H(h(x),t),t)$. The middle line is the unique map that makes the diagram commute. If you like, write $M_{\operatorname{id}_Y}$ for $Y\times I$ in the middle entry. $\endgroup$
    – Sebastian Goette
    Sep 19, 2017 at 16:44
  • $\begingroup$ Sebastian Goette I understood that you actually define a map $K :X\times I \sqcup Z \longrightarrow Y\times \{ 1\}$ by $K ((x,t))=([H(h(x),t)],1)$ and $K (z)=([g(z)],1)$ if we consider the elements of Y of the form $[y]$ in the mapping cylinder $M_h$. Since $K$ is constant on each fiber of the quotient map $X\times I\sqcup Z\longrightarrow M_{f\circ h}$, so there exists unique map $G_1 :M_{f\circ h}\longrightarrow Y\times \{ 1\}$. Now we define $G:M_{f\circ h}\longrightarrow M_{h}$ by $G([x,t])=[H(h(x),t)]$ and $G([z])=[g(z)]$. $\endgroup$
    – MHenry
    Sep 21, 2017 at 4:54
  • $\begingroup$ But I don't know yet how I define a homotopy between $G\circ F$ and $id_{M_{h}}$:( $\endgroup$
    – MHenry
    Sep 21, 2017 at 5:00

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