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$?
algebraic-topology homotopy-theory higher-homotopy-groups
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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$?
algebraic-topology homotopy-theory higher-homotopy-groups
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.
â Christian Carrick
21 hours ago
add a comment |Â
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$?
algebraic-topology homotopy-theory higher-homotopy-groups
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
algebraic-topology homotopy-theory higher-homotopy-groups
asked Sep 1 at 3:52
M.Ramana
45319
45319
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.
â Christian Carrick
21 hours ago
add a comment |Â
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.
â 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.
â 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.
â Christian Carrick
21 hours ago
add a comment |Â
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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.
â Christian Carrick
21 hours ago