Mixing
Hatcher Exercise 1.2.8
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I am trying to prove the following excersise (1.2.8) from Hatcher's Algebraic Topology:
Given 2 tori $S^1times S^1$ and identifying $S^1times {x_0}$} compute the fundamental group.
My approach is the following: Consider the first torus as an identified square via the scheme $aba^{-1}b^{-1}$ and the second torus via $cdc^{-1}d^{-1}$.
To idintify the two circles is to "glue" $a$ with $c$. Here I must make 2 remarks. The first is that we could have picked $b,d$ with no substantial change to our proof. The second is that since Hatcher doesnt describe how to glue them I assume he means in the trivial way.
Therefore , having identified $a$ with $c$ with the arrow pointing the same way we get a new CW complex with scheme $bdad^{-1}b^{-1}a^{-1}$ and therefore, from Van Campen's theorem, we the fundamental group has representation $<a,b,d|bdad^{-1}b^{-1}a^{-1}>$.
I apologise for the quality of the figure:
proof-verification algebraic-topology fundamental-groups cw-complexes
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
$endgroup$
add a comment |
$begingroup$
I am trying to prove the following excersise (1.2.8) from Hatcher's Algebraic Topology:
Given 2 tori $S^1times S^1$ and identifying $S^1times {x_0}$} compute the fundamental group.
My approach is the following: Consider the first torus as an identified square via the scheme $aba^{-1}b^{-1}$ and the second torus via $cdc^{-1}d^{-1}$.
To idintify the two circles is to "glue" $a$ with $c$. Here I must make 2 remarks. The first is that we could have picked $b,d$ with no substantial change to our proof. The second is that since Hatcher doesnt describe how to glue them I assume he means in the trivial way.
Therefore , having identified $a$ with $c$ with the arrow pointing the same way we get a new CW complex with scheme $bdad^{-1}b^{-1}a^{-1}$ and therefore, from Van Campen's theorem, we the fundamental group has representation $<a,b,d|bdad^{-1}b^{-1}a^{-1}>$.
I apologise for the quality of the figure:
proof-verification algebraic-topology fundamental-groups cw-complexes
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
$endgroup$
$begingroup$
@DietrichBurde Yes , of course , I will fix the typo.
$endgroup$
– Nick A.
Jan 28 at 13:10
$begingroup$
I think that $dfrac{F{a,b,d}}{<<bdad^{-1}b^{-1}a^{-1}>>}=dfrac{F{a,b,d}}{<<[bd,a]>>}=dfrac{F{a,b,q}}{<<[q,a]>>}$ where $q=bd$, but I'm not sure that's helpful
$endgroup$
– giannispapav
Jan 28 at 14:27
add a comment |
$begingroup$
I am trying to prove the following excersise (1.2.8) from Hatcher's Algebraic Topology:
Given 2 tori $S^1times S^1$ and identifying $S^1times {x_0}$} compute the fundamental group.
My approach is the following: Consider the first torus as an identified square via the scheme $aba^{-1}b^{-1}$ and the second torus via $cdc^{-1}d^{-1}$.
To idintify the two circles is to "glue" $a$ with $c$. Here I must make 2 remarks. The first is that we could have picked $b,d$ with no substantial change to our proof. The second is that since Hatcher doesnt describe how to glue them I assume he means in the trivial way.
Therefore , having identified $a$ with $c$ with the arrow pointing the same way we get a new CW complex with scheme $bdad^{-1}b^{-1}a^{-1}$ and therefore, from Van Campen's theorem, we the fundamental group has representation $<a,b,d|bdad^{-1}b^{-1}a^{-1}>$.
I apologise for the quality of the figure:
proof-verification algebraic-topology fundamental-groups cw-complexes
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
$endgroup$
I am trying to prove the following excersise (1.2.8) from Hatcher's Algebraic Topology:
Given 2 tori $S^1times S^1$ and identifying $S^1times {x_0}$} compute the fundamental group.
My approach is the following: Consider the first torus as an identified square via the scheme $aba^{-1}b^{-1}$ and the second torus via $cdc^{-1}d^{-1}$.
To idintify the two circles is to "glue" $a$ with $c$. Here I must make 2 remarks. The first is that we could have picked $b,d$ with no substantial change to our proof. The second is that since Hatcher doesnt describe how to glue them I assume he means in the trivial way.
Therefore , having identified $a$ with $c$ with the arrow pointing the same way we get a new CW complex with scheme $bdad^{-1}b^{-1}a^{-1}$ and therefore, from Van Campen's theorem, we the fundamental group has representation $<a,b,d|bdad^{-1}b^{-1}a^{-1}>$.
I apologise for the quality of the figure:
proof-verification algebraic-topology fundamental-groups cw-complexes
proof-verification algebraic-topology fundamental-groups cw-complexes
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
edited Jan 28 at 13:10
Nick A.
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
asked Jan 28 at 12:17
Nick A.Nick A.
1,4311320
1,4311320
$begingroup$
@DietrichBurde Yes , of course , I will fix the typo.
$endgroup$
– Nick A.
Jan 28 at 13:10
$begingroup$
I think that $dfrac{F{a,b,d}}{<<bdad^{-1}b^{-1}a^{-1}>>}=dfrac{F{a,b,d}}{<<[bd,a]>>}=dfrac{F{a,b,q}}{<<[q,a]>>}$ where $q=bd$, but I'm not sure that's helpful
$endgroup$
– giannispapav
Jan 28 at 14:27
add a comment |
$begingroup$
@DietrichBurde Yes , of course , I will fix the typo.
$endgroup$
– Nick A.
Jan 28 at 13:10
$begingroup$
I think that $dfrac{F{a,b,d}}{<<bdad^{-1}b^{-1}a^{-1}>>}=dfrac{F{a,b,d}}{<<[bd,a]>>}=dfrac{F{a,b,q}}{<<[q,a]>>}$ where $q=bd$, but I'm not sure that's helpful
$endgroup$
– giannispapav
Jan 28 at 14:27
$begingroup$
@DietrichBurde Yes , of course , I will fix the typo.
$endgroup$
– Nick A.
Jan 28 at 13:10
$begingroup$
@DietrichBurde Yes , of course , I will fix the typo.
$endgroup$
– Nick A.
Jan 28 at 13:10
$begingroup$
I think that $dfrac{F{a,b,d}}{<<bdad^{-1}b^{-1}a^{-1}>>}=dfrac{F{a,b,d}}{<<[bd,a]>>}=dfrac{F{a,b,q}}{<<[q,a]>>}$ where $q=bd$, but I'm not sure that's helpful
$endgroup$
– giannispapav
Jan 28 at 14:27
$begingroup$
I think that $dfrac{F{a,b,d}}{<<bdad^{-1}b^{-1}a^{-1}>>}=dfrac{F{a,b,d}}{<<[bd,a]>>}=dfrac{F{a,b,q}}{<<[q,a]>>}$ where $q=bd$, but I'm not sure that's helpful
$endgroup$
– giannispapav
Jan 28 at 14:27
add a comment |
1 Answer
1
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oldest
votes
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hint: can you use your figure to see that $X= (S^1 vee S^1) times S^1$?
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint: can you use your figure to see that $X= (S^1 vee S^1) times S^1$?
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
add a comment |
$begingroup$
hint: can you use your figure to see that $X= (S^1 vee S^1) times S^1$?
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
add a comment |
$begingroup$
hint: can you use your figure to see that $X= (S^1 vee S^1) times S^1$?
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
hint: can you use your figure to see that $X= (S^1 vee S^1) times S^1$?
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
answered Jan 28 at 13:42
Andres MejiaAndres Mejia
16.2k21549
16.2k21549
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
add a comment |
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
there is a similar approach to yours here
$endgroup$
– Andres Mejia
Jan 28 at 13:43
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Yes, now that you point that it is obvious! Thanks alot.
$endgroup$
– Nick A.
Jan 28 at 15:21
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
$begingroup$
Although, I mainly ask for a verification of my proof so that's why I don't accept (yet) your answer :)
$endgroup$
– Nick A.
Jan 28 at 15:25
add a comment |
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$begingroup$
@DietrichBurde Yes , of course , I will fix the typo.
$endgroup$
– Nick A.
Jan 28 at 13:10
$begingroup$
I think that $dfrac{F{a,b,d}}{<<bdad^{-1}b^{-1}a^{-1}>>}=dfrac{F{a,b,d}}{<<[bd,a]>>}=dfrac{F{a,b,q}}{<<[q,a]>>}$ where $q=bd$, but I'm not sure that's helpful
$endgroup$
– giannispapav
Jan 28 at 14:27