The effect of attaching the Möbius strip to the torusVisualising $mathbb CP^2$: a problem of attaching cells...
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The effect of attaching the Möbius strip to the torus
Visualising $mathbb CP^2$: a problem of attaching cells with a dimension gap >1Understanding attaching spaceThe First Homology Group Obtained by Attaching a Möbius Strip to a Torus in a Certain Way.Obtaining the Möbius strip as a quotient of $S^1times[-1,1]$Is there a map from the torus to the genus 2 surface which is injective on homology?Homology group of Klein bottle via Mayer Vietoris. Explanation of “ Since the boundary circle of a Möbius band wraps twice around the core circle ”Computing $pi _1(T_0cup _{S^1} M)$Fundamental group and the universal covering space for $X$ which is obtained by attaching a Mobius band to a torus.CW complex for Möbius stripCan the (extended) Möbius strip be found as the torus $T^2$ minus some embedded $S^1$?
$begingroup$
We can attach a Möbius strip $M$ to a torus by using a homeomorphism between its boundary circle and $S^1 times {x_0}$. Then the claim is that the inclusion map will send the generator of $H_1(S^1 times {x_0})$ to twice the generator of $H_1(M)$. Why is this true? Isn't $S_1 times {x_0}$ identified with the boundary
circle through a homeomorphism？ How could it wrap around the boundary circle of the Möbius strip twice then?
algebraic-topology geometric-topology
edited Mar 13 at 15:52
Bernard
123k741117
asked Mar 13 at 15:47
AlexAlex
546
$endgroup$
add a comment |
$begingroup$
We can attach a Möbius strip $M$ to a torus by using a homeomorphism between its boundary circle and $S^1 times {x_0}$. Then the claim is that the inclusion map will send the generator of $H_1(S^1 times {x_0})$ to twice the generator of $H_1(M)$. Why is this true? Isn't $S_1 times {x_0}$ identified with the boundary
circle through a homeomorphism？ How could it wrap around the boundary circle of the Möbius strip twice then?
algebraic-topology geometric-topology
edited Mar 13 at 15:52
Bernard
123k741117
asked Mar 13 at 15:47
AlexAlex
546
$endgroup$
add a comment |
$begingroup$
We can attach a Möbius strip $M$ to a torus by using a homeomorphism between its boundary circle and $S^1 times {x_0}$. Then the claim is that the inclusion map will send the generator of $H_1(S^1 times {x_0})$ to twice the generator of $H_1(M)$. Why is this true? Isn't $S_1 times {x_0}$ identified with the boundary
circle through a homeomorphism？ How could it wrap around the boundary circle of the Möbius strip twice then?
algebraic-topology geometric-topology
edited Mar 13 at 15:52
Bernard
123k741117
asked Mar 13 at 15:47
AlexAlex
546
$endgroup$
We can attach a Möbius strip $M$ to a torus by using a homeomorphism between its boundary circle and $S^1 times {x_0}$. Then the claim is that the inclusion map will send the generator of $H_1(S^1 times {x_0})$ to twice the generator of $H_1(M)$. Why is this true? Isn't $S_1 times {x_0}$ identified with the boundary
circle through a homeomorphism？ How could it wrap around the boundary circle of the Möbius strip twice then?
algebraic-topology geometric-topology
algebraic-topology geometric-topology
edited Mar 13 at 15:52
Bernard
123k741117
asked Mar 13 at 15:47
AlexAlex
546
edited Mar 13 at 15:52
Bernard
123k741117
asked Mar 13 at 15:47
AlexAlex
546
edited Mar 13 at 15:52
Bernard
123k741117
edited Mar 13 at 15:52
Bernard
123k741117
edited Mar 13 at 15:52
Bernard
123k741117
123k741117
asked Mar 13 at 15:47
AlexAlex
546
asked Mar 13 at 15:47
AlexAlex
546
asked Mar 13 at 15:47
AlexAlex
546
546
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
It does only wrap around the boundary circle once. The problem is that the boundary circle is not the generator of $H_1(M)$, it is twice the generator. Think of the deformation retract onto the middle circle to see this.
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
$endgroup$
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It does only wrap around the boundary circle once. The problem is that the boundary circle is not the generator of $H_1(M)$, it is twice the generator. Think of the deformation retract onto the middle circle to see this.
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
$endgroup$
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
add a comment |
$begingroup$
It does only wrap around the boundary circle once. The problem is that the boundary circle is not the generator of $H_1(M)$, it is twice the generator. Think of the deformation retract onto the middle circle to see this.
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
$endgroup$
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
add a comment |
$begingroup$
It does only wrap around the boundary circle once. The problem is that the boundary circle is not the generator of $H_1(M)$, it is twice the generator. Think of the deformation retract onto the middle circle to see this.
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
$endgroup$
It does only wrap around the boundary circle once. The problem is that the boundary circle is not the generator of $H_1(M)$, it is twice the generator. Think of the deformation retract onto the middle circle to see this.
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
answered Mar 13 at 15:51
Noah RiggenbachNoah Riggenbach
73528
73528
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
add a comment |
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
$begingroup$
Thank you very much that really solves my confusion!
$endgroup$
– Alex
Mar 13 at 15:53
add a comment |
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