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‎For a given map $\phi‎ :‎X\longrightarrow Y$‎, ‎the mapping cylinder of $\phi$ is defined by $M_{\phi}:=Y\bigcup_{\phi} (X \times \{ 1\})$‎. ‎Denote $\pi_n (M_{\phi},X \times \{ 1\} )$ by $\pi_n (\phi)$‎. ‎The map $\phi$ is called $n$-connected if $X$ and $Y$ are connected and $\pi_i (\phi)=0$ for $1\leq i\leq n$‎.

‎Let $X$ be a CW-complex‎. ‎Then conditions $\mathcal{F}_i$ and $\mathcal{D}_i$ on $X$ are defined inductively as follows‎:

‎$\mathcal{F}_1$‎: ‎the group $\pi_1 (X)$ is finitely generated‎.

‎$\mathcal{F}_2$‎: ‎the group $\pi_1 (X)$ is finitely presented‎, ‎and for any 1-dimensional finite CW-complex $K$ and any map $\phi‎ :‎K\longrightarrow X$ inducing an isomorphism of fundamental groups‎, ‎$\pi_2 (\phi )$ is a finitely generated module over $\mathbb{Z}\pi_1 (X)$‎.

‎$\mathcal{F}_n$‎: ‎the condition $\mathcal{F}_{n-1}$ holds‎, ‎and for any $(n-1)$-dimensional finite CW-complex $K$ and any $(n-1)$-connected map $\phi‎ :‎K\longrightarrow X$‎, ‎$\pi_n (\phi )$ is a finitely generated $\mathbb{Z}\pi_1 (X)$-module‎.

‎$\mathcal{D}_n$‎: ‎$H_i(\tilde{X})=0$ for $i>n$‎, ‎and $H^{n+1}(X;\mathcal{B})=0$ for all coefficient bundles $\mathcal{B}$.

Proposition 3.3 in th paper "Finiteness conditions for CW-complexes" of C.T.C. Wall states that : If CW-complex $X$ satisfies the conditions $\mathcal{D}_2$ and $\mathcal{F}_2$, and $\pi_1 (X)$ is free, then $X$ has the homotopy of a finite bouquet of 1-spheres and 2-spheres.

Clearly, there exist a finite bouquet of circles $K$ and a map $\phi :K\longrightarrow X$ inducing an isomorphism of fundamental groups.

Wall proved that $\pi_2 (\phi)$ is a free $\mathbb{Z}\pi_1 (X)$-module. So we can attach a finite 2-cells to $K$, necessarily with trivial attaching map, to make a new complex $L$. Since $\pi_2 (\phi)$ is free, we can extend the map $\phi$ to a $2$-connected map $\psi :L\longrightarrow X$.

My question is that:

Why $\psi$ is a homotopy equivalence? Equivalently, why does $\psi$ is i-connected for all $i\geq 3$?

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  • $\begingroup$ Surely there must also be a hypothesis like `$X$ is 2-dimensional'? The 3-sphere $S^3$ (whose fundamental group is free of rank zero) seems to give a counterexample otherwise. $\endgroup$
    – IJL
    Commented Sep 4, 2018 at 10:37
  • $\begingroup$ @IJL There is no extra hypothesis. Since there exists the map $\phi :K\longrightarrow X$ inducing ismomorphism of fundaental groups and $X$ has the condition $\mathcal{F}_2$, so $\pi_2 (\phi)$ is a finitely generated $\mathbb{Z}\pi_1 (X)$-module. Since $X$ has the condition $\mathcal{D}_2$, $\pi_2 (\phi)$ is a finitely generated projective $\mathbb{Z}\pi_1 (X)$-module (by a Lemma in the Wall's paper). Now since $\pi_1 (X)$ is free, by a well-known fact, $\pi_2 (\phi)$ is a free $\mathbb{Z}\pi_1 (X)$-module. That's why we can extend $\phi$ to a 2-connected map $\psi:L\longrightarrow X$. $\endgroup$
    – M.Ramana
    Commented Sep 4, 2018 at 12:49
  • $\begingroup$ But I don't know exactly why $\psi$ is $i$-connected for all $i$. $\endgroup$
    – M.Ramana
    Commented Sep 4, 2018 at 12:50
  • $\begingroup$ That is exactly my point: if $X=S^3$, then for any 2-complex $L$, there is only one homotopy class of maps from $L$ to $X$, and so the mapping cylinder of any $\psi:L\rightarrow X$ must be homotopy equivalent to $X=S^3$, and so no such $\psi$ can be 3-connected. There must be an extra hypothesis that $X$ is 2-dimensional, otherwise the result is not true. $\endgroup$
    – IJL
    Commented Sep 4, 2018 at 14:13
  • $\begingroup$ @IJL I think $S^3$ does not satisfy the condition $\mathcal{D}_2$. Is $H^3 (S^3 ,\mathcal{B})=0$ for all coefficient bundle $\mathcal{B}$? $\endgroup$
    – M.Ramana
    Commented Sep 5, 2018 at 7:02

1 Answer 1

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The given conditions imply that the map $\psi:L\rightarrow X$ induces an isomorphism of fundamental groups, and so it lifts to a map $\tilde\psi: \widetilde L\rightarrow \widetilde X$. The conditions also imply that $\tilde\psi$ is an isomorphism on homology. Since these spaces are simply-connected, their homology groups being isomorphic implies that their homotopy groups are isomorphic. But for $i>1$, $\pi_i(X)=\pi_i(\widetilde X)$ and similarly for $L$. Hence $\psi$ is $i$-connected for all $i$.

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  • $\begingroup$ Why the conditions imply that $\tilde{\psi}$ induces isomorphisms on homology groups? $\endgroup$
    – M.Ramana
    Commented Sep 6, 2018 at 2:47
  • $\begingroup$ Because $\pi_2(\widetilde X)$ is a free module for $\pi_1(K)\cong\pi_1(X)$, you choose the correct quantity of 2-cells to attach to $K$ to produce $L$. E.g., if $\pi_2(X)$ is free of rank 2, then you attach two 2-cells to $K$. In this way you ensure that $\pi_2(L)\cong \pi_2(X)$ as modules for $\pi_1(L)=\pi_1(K)\cong \pi_1(X)$. At this point you still have not defined the map $\psi$ on the 2-cells of $L$; you choose $\psi$ to realize some isomorphism between $\pi_2(L)$ and $\pi_2(X)$. By construction $\tilde\psi: \pi_2(\widetilde L)\rightarrow \pi_2(\widetilde X)$ is isomorphic. $\endgroup$
    – IJL
    Commented Sep 6, 2018 at 12:43
  • $\begingroup$ I understood that $\pi_2 (\psi):\pi_2 (L)\longrightarrow \pi_2 (X)$ is an ismomoephism since $\pi_2 (\tilde{L})\cong \pi_2 (L)$ and similarly for $X$. But by the Hurewitz Theorem we have only $H_2 (\tilde{X})\cong \pi_2 (\tilde{X})$ not for all $i>3$. Could you explain, for example, how to show $\psi$ is 3-connected? $\endgroup$
    – M.Ramana
    Commented Sep 7, 2018 at 3:44
  • $\begingroup$ I think we can use this version: If $f:X\to Y$ is a map of CW-complexes induces isomorphisms $\pi_1 (f):\pi_1 (X)\to \pi_1 (Y)$ and $h_i (\tilde{f}):H_i (\tilde{X})\to H_i (Y)$ for all $i>1$, then $f$ is a homotopy equivalence. Because $\psi$ induces isomorphism on fundamental groups and $H_2 (\tilde{\psi})$ is an isomorphism. On the other hand, by condition $\mathcal{D}_2$, $H_i (\tilde{X})=0$ for all i>2. Also, since $\tilde{L}$ is 2-dimensional, so $H_i (\tilde{L})=0$ for all i>2. Hence $H_i (\tilde{\psi})$ is trivially isomorphism, for all i>2. Is it true? $\endgroup$
    – M.Ramana
    Commented Sep 7, 2018 at 3:54
  • $\begingroup$ Yes, you've got it. $\endgroup$
    – IJL
    Commented Sep 7, 2018 at 9:45

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