Second homotopy group of a finite wedge of some spheres

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Let $X=bigvee S^i$ be a finite wedge of spheres containing some circles.

Why does $pi_2 (X)$ is a free $mathbbZpi_1 (X)$-module?

algebraic-topology homotopy-theory higher-homotopy-groups

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asked Jul 30 at 12:59

MHenry

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Let $X=bigvee S^i$ be a finite wedge of spheres containing some circles.

Why does $pi_2 (X)$ is a free $mathbbZpi_1 (X)$-module?

algebraic-topology homotopy-theory higher-homotopy-groups

share|cite|improve this question

asked Jul 30 at 12:59

MHenry

313

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $X=bigvee S^i$ be a finite wedge of spheres containing some circles.

Why does $pi_2 (X)$ is a free $mathbbZpi_1 (X)$-module?

algebraic-topology homotopy-theory higher-homotopy-groups

share|cite|improve this question

asked Jul 30 at 12:59

MHenry

313

Let $X=bigvee S^i$ be a finite wedge of spheres containing some circles.

Why does $pi_2 (X)$ is a free $mathbbZpi_1 (X)$-module?

algebraic-topology homotopy-theory higher-homotopy-groups

share|cite|improve this question

asked Jul 30 at 12:59

MHenry

313

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 12:59

MHenry

313

asked Jul 30 at 12:59

MHenry

313

asked Jul 30 at 12:59

MHenry

313

313

add a comment | 

add a comment | 

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Take a CW-structure on this wedge $X$ with 1 cell for each sphere. Then $pi_1$ acts on universal cover $tilde X$ freely, permuting its cells; because all cell attaching maps are nullhomotopic, $pi_2(tilde X) cong H_2(tilde X, Bbb Z)$ is a free $Bbb Z$-module generated by cells, and as action of fundamental group is free, $pi_2$ is free $Bbb Zpi_1$-module as well.

(This argument generalizes to situation when $X'$ is some complex with $Bbb Zpi_1$-free $pi_n$, and $X = X' vee S^n$.)

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answered Jul 30 at 17:25

xsnl

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1 Answer
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1 Answer
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up vote
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Take a CW-structure on this wedge $X$ with 1 cell for each sphere. Then $pi_1$ acts on universal cover $tilde X$ freely, permuting its cells; because all cell attaching maps are nullhomotopic, $pi_2(tilde X) cong H_2(tilde X, Bbb Z)$ is a free $Bbb Z$-module generated by cells, and as action of fundamental group is free, $pi_2$ is free $Bbb Zpi_1$-module as well.

(This argument generalizes to situation when $X'$ is some complex with $Bbb Zpi_1$-free $pi_n$, and $X = X' vee S^n$.)

share|cite|improve this answer

answered Jul 30 at 17:25

xsnl

1,155418

add a comment | 

up vote
1
down vote

Take a CW-structure on this wedge $X$ with 1 cell for each sphere. Then $pi_1$ acts on universal cover $tilde X$ freely, permuting its cells; because all cell attaching maps are nullhomotopic, $pi_2(tilde X) cong H_2(tilde X, Bbb Z)$ is a free $Bbb Z$-module generated by cells, and as action of fundamental group is free, $pi_2$ is free $Bbb Zpi_1$-module as well.

(This argument generalizes to situation when $X'$ is some complex with $Bbb Zpi_1$-free $pi_n$, and $X = X' vee S^n$.)

share|cite|improve this answer

answered Jul 30 at 17:25

xsnl

1,155418

add a comment | 

up vote
1
down vote

up vote
1
down vote

Take a CW-structure on this wedge $X$ with 1 cell for each sphere. Then $pi_1$ acts on universal cover $tilde X$ freely, permuting its cells; because all cell attaching maps are nullhomotopic, $pi_2(tilde X) cong H_2(tilde X, Bbb Z)$ is a free $Bbb Z$-module generated by cells, and as action of fundamental group is free, $pi_2$ is free $Bbb Zpi_1$-module as well.

(This argument generalizes to situation when $X'$ is some complex with $Bbb Zpi_1$-free $pi_n$, and $X = X' vee S^n$.)

share|cite|improve this answer

answered Jul 30 at 17:25

xsnl

1,155418

Take a CW-structure on this wedge $X$ with 1 cell for each sphere. Then $pi_1$ acts on universal cover $tilde X$ freely, permuting its cells; because all cell attaching maps are nullhomotopic, $pi_2(tilde X) cong H_2(tilde X, Bbb Z)$ is a free $Bbb Z$-module generated by cells, and as action of fundamental group is free, $pi_2$ is free $Bbb Zpi_1$-module as well.

(This argument generalizes to situation when $X'$ is some complex with $Bbb Zpi_1$-free $pi_n$, and $X = X' vee S^n$.)

share|cite|improve this answer

answered Jul 30 at 17:25

xsnl

1,155418

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 17:25

xsnl

1,155418

answered Jul 30 at 17:25

xsnl

1,155418

answered Jul 30 at 17:25

xsnl

1,155418

1,155418

add a comment | 

add a comment | 

 

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