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Finding the fundamental group of a subspace of $mathbbR^4$

1

$begingroup$

Let $C^times := mathbbC backslash 0$ and $Z := (w, z) in left(mathbbC^timesright)^2 $ for some fixed $n geq 1$. I'm trying to find the fundamental group of $Z$. My toolbox consists pretty much out of Seifert-van Kampen and deformation retracting to a CW-complex, but I don't know how to do that here because $Z$ is 4 dimensional. Here is my work so far:

For a given $z$, the solutions are $w_1, ..., w_n$ where $w_i$ has length $sqrt[n]w$ and argument $frac2in cdotpi $, and the fundamental group of $mathbbC^times$ is $mathbbZ$ because it deformation retracts unto $ = 1$.

algebraic-topology complex-numbers

share|cite|improve this question

edited Mar 30 at 16:19

Pel de Pinda

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The space $Z$ is not $4$-dimensional. Regardless, it is a cover of $C^*$, and it follows that its fundamental group is a subgroup of that of $C^*$.
  $endgroup$
  – Amitai Yuval
  Mar 30 at 16:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I guess I meant to write covering.
  $endgroup$
  – Amitai Yuval
  Mar 30 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmitaiYuval Thanks, I should have said subset of $mathbbR^4$. I know it is a covering of $mathbbC^times$, I wanted to find the image of the induced homomorphism $p_*: pi_1(Z, z_0) to pi_1(mathbbC^times, c_0)$ by finding a generator for $pi_1(Z, z_0)$ and then seeing where $p$ maps it to. Is there a way to do this without finding the fundamental group of $Z$ explicitely?
  $endgroup$
  – Pel de Pinda
  Mar 30 at 16:19
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $C^times := mathbbC backslash 0$ and $Z := (w, z) in left(mathbbC^timesright)^2 $ for some fixed $n geq 1$. I'm trying to find the fundamental group of $Z$. My toolbox consists pretty much out of Seifert-van Kampen and deformation retracting to a CW-complex, but I don't know how to do that here because $Z$ is 4 dimensional. Here is my work so far:

For a given $z$, the solutions are $w_1, ..., w_n$ where $w_i$ has length $sqrt[n]w$ and argument $frac2in cdotpi $, and the fundamental group of $mathbbC^times$ is $mathbbZ$ because it deformation retracts unto $ = 1$.

algebraic-topology complex-numbers

share|cite|improve this question

edited Mar 30 at 16:19

Pel de Pinda

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The space $Z$ is not $4$-dimensional. Regardless, it is a cover of $C^*$, and it follows that its fundamental group is a subgroup of that of $C^*$.
  $endgroup$
  – Amitai Yuval
  Mar 30 at 16:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I guess I meant to write covering.
  $endgroup$
  – Amitai Yuval
  Mar 30 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmitaiYuval Thanks, I should have said subset of $mathbbR^4$. I know it is a covering of $mathbbC^times$, I wanted to find the image of the induced homomorphism $p_*: pi_1(Z, z_0) to pi_1(mathbbC^times, c_0)$ by finding a generator for $pi_1(Z, z_0)$ and then seeing where $p$ maps it to. Is there a way to do this without finding the fundamental group of $Z$ explicitely?
  $endgroup$
  – Pel de Pinda
  Mar 30 at 16:19
  

  
  
  
  

add a comment | 

$begingroup$

Let $C^times := mathbbC backslash 0$ and $Z := (w, z) in left(mathbbC^timesright)^2 $ for some fixed $n geq 1$. I'm trying to find the fundamental group of $Z$. My toolbox consists pretty much out of Seifert-van Kampen and deformation retracting to a CW-complex, but I don't know how to do that here because $Z$ is 4 dimensional. Here is my work so far:

For a given $z$, the solutions are $w_1, ..., w_n$ where $w_i$ has length $sqrt[n]w$ and argument $frac2in cdotpi $, and the fundamental group of $mathbbC^times$ is $mathbbZ$ because it deformation retracts unto $ = 1$.

algebraic-topology complex-numbers

share|cite|improve this question

edited Mar 30 at 16:19

Pel de Pinda

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

$endgroup$

share|cite|improve this question

edited Mar 30 at 16:19

Pel de Pinda

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

share|cite|improve this question

edited Mar 30 at 16:19

Pel de Pinda

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

asked Mar 30 at 16:01

Pel de PindaPel de Pinda

929416

1 Answer
1

active

oldest

votes

3

$begingroup$

We can write $Z = (w, w^n) mid w in mathbbC^times $. Let $alpha : mathbbC^times to mathbbC^times, alpha(w) = w^n$. We see that $Z$ is nothing else than the graph of $alpha$.

But for any contiunuous map $phi : X to Y$ between topological spaces $X,Y$ the graph $G(phi) = (x,phi(x)) mid x in X subset X times Y$ is homeomorphic to $X$. In fact, define $f : X to G(phi), f(x) = (x,phi(x))$ and $g : G(phi) to X, g(x,phi(x)) = x$. These are continuous maps (note that $g$ is the restriction of the projection $X times Y to X$) such that $g circ f = id_X$ and $f circ g = id_G(phi)$.

Hence $Z approx mathbbC^times$ and $pi_1(Z) approx mathbbZ$.

share|cite|improve this answer

edited Mar 30 at 16:52

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035

$endgroup$

add a comment | 

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3

$begingroup$

We can write $Z = (w, w^n) mid w in mathbbC^times $. Let $alpha : mathbbC^times to mathbbC^times, alpha(w) = w^n$. We see that $Z$ is nothing else than the graph of $alpha$.

But for any contiunuous map $phi : X to Y$ between topological spaces $X,Y$ the graph $G(phi) = (x,phi(x)) mid x in X subset X times Y$ is homeomorphic to $X$. In fact, define $f : X to G(phi), f(x) = (x,phi(x))$ and $g : G(phi) to X, g(x,phi(x)) = x$. These are continuous maps (note that $g$ is the restriction of the projection $X times Y to X$) such that $g circ f = id_X$ and $f circ g = id_G(phi)$.

Hence $Z approx mathbbC^times$ and $pi_1(Z) approx mathbbZ$.

share|cite|improve this answer

edited Mar 30 at 16:52

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035

$endgroup$

add a comment | 

3

$begingroup$

We can write $Z = (w, w^n) mid w in mathbbC^times $. Let $alpha : mathbbC^times to mathbbC^times, alpha(w) = w^n$. We see that $Z$ is nothing else than the graph of $alpha$.

But for any contiunuous map $phi : X to Y$ between topological spaces $X,Y$ the graph $G(phi) = (x,phi(x)) mid x in X subset X times Y$ is homeomorphic to $X$. In fact, define $f : X to G(phi), f(x) = (x,phi(x))$ and $g : G(phi) to X, g(x,phi(x)) = x$. These are continuous maps (note that $g$ is the restriction of the projection $X times Y to X$) such that $g circ f = id_X$ and $f circ g = id_G(phi)$.

Hence $Z approx mathbbC^times$ and $pi_1(Z) approx mathbbZ$.

share|cite|improve this answer

edited Mar 30 at 16:52

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035

$endgroup$

add a comment | 

$begingroup$

We can write $Z = (w, w^n) mid w in mathbbC^times $. Let $alpha : mathbbC^times to mathbbC^times, alpha(w) = w^n$. We see that $Z$ is nothing else than the graph of $alpha$.

But for any contiunuous map $phi : X to Y$ between topological spaces $X,Y$ the graph $G(phi) = (x,phi(x)) mid x in X subset X times Y$ is homeomorphic to $X$. In fact, define $f : X to G(phi), f(x) = (x,phi(x))$ and $g : G(phi) to X, g(x,phi(x)) = x$. These are continuous maps (note that $g$ is the restriction of the projection $X times Y to X$) such that $g circ f = id_X$ and $f circ g = id_G(phi)$.

Hence $Z approx mathbbC^times$ and $pi_1(Z) approx mathbbZ$.

share|cite|improve this answer

edited Mar 30 at 16:52

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035

$endgroup$

share|cite|improve this answer

edited Mar 30 at 16:52

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035

answered Mar 30 at 16:45

Paul FrostPaul Frost

12.6k31035