Suppose that $Z$ is a finite wedge of spheres containing circles and there exist maps $f:Y\to Z$ and $g:Z\to Y$ so that $g\circ f\simeq 1_Y$. Assume that there exists a map $h:X\to Y$ which induces an isomorphism on fundamental groups, where $X$ is a finite wedge of circles.

Is $\pi_2 (M_{h},X)$ a projective (or a free) $\mathbb{Z}\pi_1 (X)$-module?

Here $M_h =\frac{X\times I\sqcup Y}{(x,1)\sim h(x)}$ denotes the mapping cylinder of $h$.

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    $\begingroup$ How is $\pi_2(M_h,X)$ a module over $\mathbb{Z}[\pi_1(X)]$ at all? Relative $\pi_2$ is not even an abelian group in general, and I don't think that the maps go the right way to give it an action of $\pi_1(X)$. $\endgroup$ – Neil Strickland Jul 3 '18 at 18:47
  • $\begingroup$ @NeilStrickland Yes, but here since $h$ induces an isomorphism on fundamental groups and $\pi_2 (X)=0$, we have $\pi_2 (M_h ,X)\cong \pi_2 (Y)$. Hence $\pi_2 (M_h ,X)$ is abelian. $\endgroup$ – MHenry Jul 4 '18 at 4:18
  • $\begingroup$ @NeilStrickland By [Hatcher, p. 345], $\pi_1 (X)$ acts on the whole long exact sequence of homotopy groups for $(M_{h},X)$, the action commuting with the various maps in the sequence. $\endgroup$ – MHenry Jul 4 '18 at 4:23

Based on your comments to this answer, you seem to be in the situation where $\pi_2(Z)$ is a free $\mathbb{Z}\pi_1(X)$-module, and where the map $\pi_1(X)\to \pi_1(Y)$ is an iso. Under those assumptions, the answer should be a consequence of the following diagram, coming from the map between the mapping cylinders :

$$\require{AMScd}\begin{CD} 0@>>>\pi_2(Y)@>\varphi_3>>\pi_2(M_h,X)@>\varphi_2=0>>\pi_1(X)@>\varphi_1>>\pi_1(Y)\\ @VVV@VVV@VVV@VVV@VVV\\ 0@>>>\pi_2(Z)@>>>\pi_2(M_{f\circ h},X)@>\varphi'_2>>\pi_1(X)@>>>\pi_1(Z)\\ @VVV@VVV@VVV@VVV@VVV\\ 0@>>>\pi_2(Y)@>\varphi_3>>\pi_2(M_h,X)@>\varphi_2=0>>\pi_1(X)@>\varphi_1>>\pi_1(Y)\\ \end{CD}$$

Now, by assumption, $\varphi_1$ is an iso. This implies that its kernel is $0$, which means that $\varphi_2=0$. This means in turn that $\varphi_3$ is an iso. But then, by assumption $\pi_2(Z)$ is a free $\mathbb{Z}\pi_1(X)$-module, and so $\pi_2(M_h,X)\simeq\pi_2(Y)$ is a projective $\mathbb{Z}\pi_1(X)$-module since it is a direct summand of $\pi_2(Z)$.

In addition, using the commutativity of the diagram, we get that $\varphi'_2=0$, which means that $\pi_2(M_{f\circ h},X)$ is isomorphic to $\pi_2(Z)$, and in particular it is a free $\mathbb{Z}\pi_1(X)$-module.

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  • $\begingroup$ Thank you so much for your answer and your help. I've learned from you a lot. Thanks. $\endgroup$ – MHenry Jul 20 '18 at 4:21
  • $\begingroup$ I was wondering if I could you ask another question about the above problem. Why $\pi_2 (Z)$ is a free $\mathbb{Z}\pi_1 (Z)$-module? $\endgroup$ – MHenry Aug 31 '18 at 4:00
  • $\begingroup$ I am not sure why $\pi_2(Z)$ is a free $\mathbb{Z}\pi_1(X)$-module. The proof I wrote relies only on the long exact sequences and the map between them, and I used the assumptions you wrote in your question and in the comments here $\endgroup$ – S. Douteau Sep 7 '18 at 8:48

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