Second homotopy group of a special mapping cylinder

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Let $X$ be a finite wedge of spheres containing some circles and $A$ is homotopy dominated by $X$, i.e. there exist $f:Alongrightarrow X$ and $g:Xlongrightarrow A$ so that $gcirc fsimeq 1_A$. Suppose that the map $phi :Klongrightarrow A$ induces an isomorphism of fundamental groups, where $K$ is a wedge of circles.



If $M_phi$ denotes the mapping cylinder of $phi$, then is $pi_2 (M_phi,K)$ a free $mathbbZpi_1 (K)$-module$?










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  • How about if $X=A=K$ and all maps are the identity, then the group in question vanishes, and so can't be a free module.
    – Christian Carrick
    21 hours ago














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Let $X$ be a finite wedge of spheres containing some circles and $A$ is homotopy dominated by $X$, i.e. there exist $f:Alongrightarrow X$ and $g:Xlongrightarrow A$ so that $gcirc fsimeq 1_A$. Suppose that the map $phi :Klongrightarrow A$ induces an isomorphism of fundamental groups, where $K$ is a wedge of circles.



If $M_phi$ denotes the mapping cylinder of $phi$, then is $pi_2 (M_phi,K)$ a free $mathbbZpi_1 (K)$-module$?










share|cite|improve this question





















  • How about if $X=A=K$ and all maps are the identity, then the group in question vanishes, and so can't be a free module.
    – Christian Carrick
    21 hours ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $X$ be a finite wedge of spheres containing some circles and $A$ is homotopy dominated by $X$, i.e. there exist $f:Alongrightarrow X$ and $g:Xlongrightarrow A$ so that $gcirc fsimeq 1_A$. Suppose that the map $phi :Klongrightarrow A$ induces an isomorphism of fundamental groups, where $K$ is a wedge of circles.



If $M_phi$ denotes the mapping cylinder of $phi$, then is $pi_2 (M_phi,K)$ a free $mathbbZpi_1 (K)$-module$?










share|cite|improve this question













Let $X$ be a finite wedge of spheres containing some circles and $A$ is homotopy dominated by $X$, i.e. there exist $f:Alongrightarrow X$ and $g:Xlongrightarrow A$ so that $gcirc fsimeq 1_A$. Suppose that the map $phi :Klongrightarrow A$ induces an isomorphism of fundamental groups, where $K$ is a wedge of circles.



If $M_phi$ denotes the mapping cylinder of $phi$, then is $pi_2 (M_phi,K)$ a free $mathbbZpi_1 (K)$-module$?







algebraic-topology homotopy-theory higher-homotopy-groups






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asked Sep 1 at 3:52









M.Ramana

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  • How about if $X=A=K$ and all maps are the identity, then the group in question vanishes, and so can't be a free module.
    – Christian Carrick
    21 hours ago
















  • How about if $X=A=K$ and all maps are the identity, then the group in question vanishes, and so can't be a free module.
    – Christian Carrick
    21 hours ago















How about if $X=A=K$ and all maps are the identity, then the group in question vanishes, and so can't be a free module.
– Christian Carrick
21 hours ago




How about if $X=A=K$ and all maps are the identity, then the group in question vanishes, and so can't be a free module.
– Christian Carrick
21 hours ago















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