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Construction of a K(pi_1)- space?

Does a function space construction always decrease the connectivity of a space?$pi_1(S^n)=0$ for $ngeq2$Question on part of the proof that $pi_1(X,x_0)$ is isomorphic to $pi_1(X,x_1)$A standard proof of $pi_1(mathbbS^1) = mathbbZ$ using universal covering spacesRole of the Thom space in the Pontryagin-Thom constructionShow $pi_1(SU(2)/H_8)$Construction of K(G,1) spaceHomotopy constructionProve that $pi_1(Xtimes Y, (x_0,y_0))$ is isomorphic to $pi_1(X,x_0)times pi_1(Y,y_0)$.The meaning of $pi_1(S^1,(1,0))$

1

1

$begingroup$

I was advised to post my question here. A colleague suggested a proof of a fact which I have hard time to believe. Since I am not a topologist by training I wonder if this can be true in such a generality.

Consider any finite CW-complex structure on the 𝑑-dimensional sphere $𝑆^𝑑$. Let $𝐾$ be $𝑆^𝑑$ with all strata of codimension at least 2 removed, i.e. we keep just 𝑑 and (𝑑−1)-dimensional strata of the complex under consideration.

Claim. K is contractible to a graph (the dual graph of the preserved strata). Thus it is a $K(pi_1, 1)$- space where $pi_1$ is a free group.

The claim is trivially true for d=2, but looks suspicious for higher d. Also if it is true then it might work for more general CW-complexes as well.

Best regards, Boris Shapiro/Stockholm University/

homotopy-theory

share|cite|improve this question

asked Mar 21 at 6:30

user53092user53092

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean more precisely ? For instance for $d=2$ if you take the standard cell-structure with one $2$-cell and one $0$-cell, how do you un-collapse the boundary of the $2$-cell ?
  $endgroup$
  – Max
  Mar 21 at 8:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, in this case you remove the point. The remaining part in the 2-disk which is contractible to a point. $pi_1$ is trivial. In general, in 2 dimensions, you remove a bunch of points (at least one). The remaining punched disc is homotopy equivalent to a wedge of circles.
  $endgroup$
  – user53092
  Mar 21 at 10:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh ok so just literally remove. Sorry for my question, I thought you meant something more sophisticated.
  $endgroup$
  – Max
  Mar 21 at 10:58
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I was advised to post my question here. A colleague suggested a proof of a fact which I have hard time to believe. Since I am not a topologist by training I wonder if this can be true in such a generality.

Consider any finite CW-complex structure on the 𝑑-dimensional sphere $𝑆^𝑑$. Let $𝐾$ be $𝑆^𝑑$ with all strata of codimension at least 2 removed, i.e. we keep just 𝑑 and (𝑑−1)-dimensional strata of the complex under consideration.

Claim. K is contractible to a graph (the dual graph of the preserved strata). Thus it is a $K(pi_1, 1)$- space where $pi_1$ is a free group.

The claim is trivially true for d=2, but looks suspicious for higher d. Also if it is true then it might work for more general CW-complexes as well.

Best regards, Boris Shapiro/Stockholm University/

homotopy-theory

share|cite|improve this question

asked Mar 21 at 6:30

user53092user53092

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean more precisely ? For instance for $d=2$ if you take the standard cell-structure with one $2$-cell and one $0$-cell, how do you un-collapse the boundary of the $2$-cell ?
  $endgroup$
  – Max
  Mar 21 at 8:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, in this case you remove the point. The remaining part in the 2-disk which is contractible to a point. $pi_1$ is trivial. In general, in 2 dimensions, you remove a bunch of points (at least one). The remaining punched disc is homotopy equivalent to a wedge of circles.
  $endgroup$
  – user53092
  Mar 21 at 10:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh ok so just literally remove. Sorry for my question, I thought you meant something more sophisticated.
  $endgroup$
  – Max
  Mar 21 at 10:58
  

  
  
  
  

add a comment | 

$begingroup$

I was advised to post my question here. A colleague suggested a proof of a fact which I have hard time to believe. Since I am not a topologist by training I wonder if this can be true in such a generality.

Consider any finite CW-complex structure on the 𝑑-dimensional sphere $𝑆^𝑑$. Let $𝐾$ be $𝑆^𝑑$ with all strata of codimension at least 2 removed, i.e. we keep just 𝑑 and (𝑑−1)-dimensional strata of the complex under consideration.

Claim. K is contractible to a graph (the dual graph of the preserved strata). Thus it is a $K(pi_1, 1)$- space where $pi_1$ is a free group.

The claim is trivially true for d=2, but looks suspicious for higher d. Also if it is true then it might work for more general CW-complexes as well.

Best regards, Boris Shapiro/Stockholm University/

homotopy-theory

share|cite|improve this question

asked Mar 21 at 6:30

user53092user53092

62

$endgroup$

share|cite|improve this question

asked Mar 21 at 6:30

user53092user53092

62

share|cite|improve this question

asked Mar 21 at 6:30

user53092user53092

62

asked Mar 21 at 6:30

user53092user53092

62

asked Mar 21 at 6:30

user53092user53092

62