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How do I go from 300 unfinished/half written blog posts, to published posts?

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Constructing a Universal Cover--Proving Injectivity

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$begingroup$

Here is a quote from Hatcher's Algebraic Topology:

Given a set $U in mathcalU$ and a path $gamma$ in $X$ from $x_0$ to a point in $U$, let $$U_[gamma] = [ gamma cdot eta ] mid eta text is a path in U text with eta (0) = gamma (1) $$ As the notation indicates, $U_[gamma]$ depends only on the homotopy class $[gamma ]$. Observe that $p : U_[gamma] to U$ is surjective since $U$ is path-connected and injective since different choices of $eta$ joining $gamma (1)$ to a fixed $x in U$ are all homotopic in $X$, the mapping $pi_1(U) to pi_1(X)$ being trivial.

See page 64 for more context. Note that $p : widetildeX to X$ is defined as $p([gamma]) = gamma (1)$, where $gamma$ is a path in $X$ starting at $x_0$. I can see that the map is surjective. However, I am having a little troubling working out the injectivity of the map.

If $iota : U to X$ denotes the canonical embedding, then the above quote says that $iota_* : pi_1 (U) to pi_1(X)$ is trivial (in fact, it is trivial for any base point). Suppose that $eta_1$ and $eta_2$ are paths in $U$ with the same starting and ending point such that $p([gamma cdot eta_1]) = p([gamma cdot eta_2])$. This implies $eta_1 (1) =eta_2 (1)$. Then $[eta_1 cdot overlineeta_2] in pi_1(U)$, so $iota_*([eta_1 cdot overlineeta_2]) = [iota circ (eta_1 cdot overlineeta_2]$ is trivial in $pi_1(X)$, which implies $iota circ eta_1 simeq iota circ eta_2$. Hatcher seems to be claiming that this implies $eta_1 simeq eta_2$. If this is what he is asserting, I don't see how to deduce that conclusion.

algebraic-topology homotopy-theory fundamental-groups

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asked Mar 17 at 21:02

user193319user193319

2,4532927

$endgroup$

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  No that's not what he was asserting at all, he did say "homotopic in $X$". So $iotacirc eta_1 simeq iotacirc eta_2 implies [gammaeta_1]=[gammaeta_2]$, where in this second equality everything is in $X$
  $endgroup$
  – Max
  Mar 17 at 21:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max Oh, okay. It was a slight abuse in notation that was confusing me.
  $endgroup$
  – user193319
  Mar 18 at 0:21
  

  
  
  
  

add a comment | 

0

$begingroup$

Here is a quote from Hatcher's Algebraic Topology:

Given a set $U in mathcalU$ and a path $gamma$ in $X$ from $x_0$ to a point in $U$, let $$U_[gamma] = [ gamma cdot eta ] mid eta text is a path in U text with eta (0) = gamma (1) $$ As the notation indicates, $U_[gamma]$ depends only on the homotopy class $[gamma ]$. Observe that $p : U_[gamma] to U$ is surjective since $U$ is path-connected and injective since different choices of $eta$ joining $gamma (1)$ to a fixed $x in U$ are all homotopic in $X$, the mapping $pi_1(U) to pi_1(X)$ being trivial.

See page 64 for more context. Note that $p : widetildeX to X$ is defined as $p([gamma]) = gamma (1)$, where $gamma$ is a path in $X$ starting at $x_0$. I can see that the map is surjective. However, I am having a little troubling working out the injectivity of the map.

If $iota : U to X$ denotes the canonical embedding, then the above quote says that $iota_* : pi_1 (U) to pi_1(X)$ is trivial (in fact, it is trivial for any base point). Suppose that $eta_1$ and $eta_2$ are paths in $U$ with the same starting and ending point such that $p([gamma cdot eta_1]) = p([gamma cdot eta_2])$. This implies $eta_1 (1) =eta_2 (1)$. Then $[eta_1 cdot overlineeta_2] in pi_1(U)$, so $iota_*([eta_1 cdot overlineeta_2]) = [iota circ (eta_1 cdot overlineeta_2]$ is trivial in $pi_1(X)$, which implies $iota circ eta_1 simeq iota circ eta_2$. Hatcher seems to be claiming that this implies $eta_1 simeq eta_2$. If this is what he is asserting, I don't see how to deduce that conclusion.

algebraic-topology homotopy-theory fundamental-groups

share|cite|improve this question

asked Mar 17 at 21:02

user193319user193319

2,4532927

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No that's not what he was asserting at all, he did say "homotopic in $X$". So $iotacirc eta_1 simeq iotacirc eta_2 implies [gammaeta_1]=[gammaeta_2]$, where in this second equality everything is in $X$
  $endgroup$
  – Max
  Mar 17 at 21:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max Oh, okay. It was a slight abuse in notation that was confusing me.
  $endgroup$
  – user193319
  Mar 18 at 0:21
  

  
  
  
  

add a comment | 

$begingroup$

Here is a quote from Hatcher's Algebraic Topology:

Given a set $U in mathcalU$ and a path $gamma$ in $X$ from $x_0$ to a point in $U$, let $$U_[gamma] = [ gamma cdot eta ] mid eta text is a path in U text with eta (0) = gamma (1) $$ As the notation indicates, $U_[gamma]$ depends only on the homotopy class $[gamma ]$. Observe that $p : U_[gamma] to U$ is surjective since $U$ is path-connected and injective since different choices of $eta$ joining $gamma (1)$ to a fixed $x in U$ are all homotopic in $X$, the mapping $pi_1(U) to pi_1(X)$ being trivial.

See page 64 for more context. Note that $p : widetildeX to X$ is defined as $p([gamma]) = gamma (1)$, where $gamma$ is a path in $X$ starting at $x_0$. I can see that the map is surjective. However, I am having a little troubling working out the injectivity of the map.

If $iota : U to X$ denotes the canonical embedding, then the above quote says that $iota_* : pi_1 (U) to pi_1(X)$ is trivial (in fact, it is trivial for any base point). Suppose that $eta_1$ and $eta_2$ are paths in $U$ with the same starting and ending point such that $p([gamma cdot eta_1]) = p([gamma cdot eta_2])$. This implies $eta_1 (1) =eta_2 (1)$. Then $[eta_1 cdot overlineeta_2] in pi_1(U)$, so $iota_*([eta_1 cdot overlineeta_2]) = [iota circ (eta_1 cdot overlineeta_2]$ is trivial in $pi_1(X)$, which implies $iota circ eta_1 simeq iota circ eta_2$. Hatcher seems to be claiming that this implies $eta_1 simeq eta_2$. If this is what he is asserting, I don't see how to deduce that conclusion.

algebraic-topology homotopy-theory fundamental-groups

share|cite|improve this question

asked Mar 17 at 21:02

user193319user193319

2,4532927

$endgroup$

share|cite|improve this question

asked Mar 17 at 21:02

user193319user193319

2,4532927

share|cite|improve this question

asked Mar 17 at 21:02

user193319user193319

2,4532927

asked Mar 17 at 21:02

user193319user193319

2,4532927

asked Mar 17 at 21:02

user193319user193319

2,4532927