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0

$begingroup$

Let $M$ be a closed orientable $n$-dimensional manifold and $Sigma$ be a $2$-dimensional sphere embedded in $M$ such that there is a map $f:Sigmarightarrow mathbbS^2$ with non zero degree, i.e., $deg (f)not=0$. Is it true that the embedding of $Sigma$ in $M$ represent a non trivial element of $pi_2(M)$? if so, how can I prove it?

algebraic-topology differential-topology

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edited Mar 22 at 17:23

Michael

asked Mar 22 at 16:27

MichaelMichael

383

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please say "second homotopy group" rather than "second fundamental group".
  $endgroup$
  – John Palmieri
  Mar 22 at 17:20
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $M$ be a closed orientable $n$-dimensional manifold and $Sigma$ be a $2$-dimensional sphere embedded in $M$ such that there is a map $f:Sigmarightarrow mathbbS^2$ with non zero degree, i.e., $deg (f)not=0$. Is it true that the embedding of $Sigma$ in $M$ represent a non trivial element of $pi_2(M)$? if so, how can I prove it?

algebraic-topology differential-topology

share|cite|improve this question

edited Mar 22 at 17:23

Michael

asked Mar 22 at 16:27

MichaelMichael

383

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please say "second homotopy group" rather than "second fundamental group".
  $endgroup$
  – John Palmieri
  Mar 22 at 17:20
  

  
  
  
  

add a comment | 

$begingroup$

Let $M$ be a closed orientable $n$-dimensional manifold and $Sigma$ be a $2$-dimensional sphere embedded in $M$ such that there is a map $f:Sigmarightarrow mathbbS^2$ with non zero degree, i.e., $deg (f)not=0$. Is it true that the embedding of $Sigma$ in $M$ represent a non trivial element of $pi_2(M)$? if so, how can I prove it?

algebraic-topology differential-topology

share|cite|improve this question

edited Mar 22 at 17:23

Michael

asked Mar 22 at 16:27

MichaelMichael

383

$endgroup$

share|cite|improve this question

edited Mar 22 at 17:23

Michael

asked Mar 22 at 16:27

MichaelMichael

383

share|cite|improve this question

edited Mar 22 at 17:23

Michael

asked Mar 22 at 16:27

MichaelMichael

383

asked Mar 22 at 16:27

MichaelMichael

383

asked Mar 22 at 16:27

MichaelMichael

383

2 Answers
2

active

oldest

votes

0

$begingroup$

Let $M$ be $S^3$, The stereotgraphic projection allows to identify $S^3-point$ to $mathbbR^3$, there are spheres imbedded in $mathbbR^3$ but $pi_2(S^3)=1$.

share|cite|improve this answer

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

add a comment | 

0

$begingroup$

I'm guess you mean that there is a map $f:Mto S^2$ whose restriction to $Sigma$ has non-zero degree. In that case it is true: otherwise there is a homotopy $F:Sigma times [0, 1] to M$ contarcting $Sigma$ to a point, and composing $F$ with $f$ gives a contracting homotopy from $f|_Sigma=fcdot F_0: Sigma to S^2$ to the constant map $fcdot F_1: Sigma to S^2$

share|cite|improve this answer

answered Mar 22 at 18:04

MaxMax

4,4421326

$endgroup$

add a comment | 

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0

$begingroup$

Let $M$ be $S^3$, The stereotgraphic projection allows to identify $S^3-point$ to $mathbbR^3$, there are spheres imbedded in $mathbbR^3$ but $pi_2(S^3)=1$.

share|cite|improve this answer

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

add a comment | 

0

$begingroup$

Let $M$ be $S^3$, The stereotgraphic projection allows to identify $S^3-point$ to $mathbbR^3$, there are spheres imbedded in $mathbbR^3$ but $pi_2(S^3)=1$.

share|cite|improve this answer

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

add a comment | 

$begingroup$

Let $M$ be $S^3$, The stereotgraphic projection allows to identify $S^3-point$ to $mathbbR^3$, there are spheres imbedded in $mathbbR^3$ but $pi_2(S^3)=1$.

share|cite|improve this answer

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

share|cite|improve this answer

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

answered Mar 22 at 16:33

Tsemo AristideTsemo Aristide

60.4k11446

0

$begingroup$

I'm guess you mean that there is a map $f:Mto S^2$ whose restriction to $Sigma$ has non-zero degree. In that case it is true: otherwise there is a homotopy $F:Sigma times [0, 1] to M$ contarcting $Sigma$ to a point, and composing $F$ with $f$ gives a contracting homotopy from $f|_Sigma=fcdot F_0: Sigma to S^2$ to the constant map $fcdot F_1: Sigma to S^2$

share|cite|improve this answer

answered Mar 22 at 18:04

MaxMax

4,4421326

$endgroup$

add a comment | 

0

$begingroup$

I'm guess you mean that there is a map $f:Mto S^2$ whose restriction to $Sigma$ has non-zero degree. In that case it is true: otherwise there is a homotopy $F:Sigma times [0, 1] to M$ contarcting $Sigma$ to a point, and composing $F$ with $f$ gives a contracting homotopy from $f|_Sigma=fcdot F_0: Sigma to S^2$ to the constant map $fcdot F_1: Sigma to S^2$

share|cite|improve this answer

answered Mar 22 at 18:04

MaxMax

4,4421326

$endgroup$

add a comment | 

$begingroup$

I'm guess you mean that there is a map $f:Mto S^2$ whose restriction to $Sigma$ has non-zero degree. In that case it is true: otherwise there is a homotopy $F:Sigma times [0, 1] to M$ contarcting $Sigma$ to a point, and composing $F$ with $f$ gives a contracting homotopy from $f|_Sigma=fcdot F_0: Sigma to S^2$ to the constant map $fcdot F_1: Sigma to S^2$

share|cite|improve this answer

answered Mar 22 at 18:04

MaxMax

4,4421326

$endgroup$

share|cite|improve this answer

answered Mar 22 at 18:04

MaxMax

4,4421326

answered Mar 22 at 18:04

MaxMax

4,4421326

answered Mar 22 at 18:04

MaxMax

4,4421326