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Problems Part 1 Let M be the Möbius Strip obtained from identifying (0, t) with (1, 1 − t) on the square I × I. Let A be the boundary circle i.e. the image of I × {0, 1} under the quotient map....

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Problems
Part 1
Let M be the Möbius Strip obtained from identifying (0, t) with (1, 1 − t) on
the square I × I. Let A be the boundary circle i.e. the image of I × {0, 1}
under the quotient map. Recall that using the long exact sequence of the pai
(M,A), the first relative homology H1(M,A) ∼= Z/2Z. Prove using the long
exact sequence of the triple (M,B,A) that the curve f(t) = (0, t) (thought of
as a 1-chain) represents the non-zero class in this group. Here B is the image
of the boundary of I × I in M .
Part 2
Let (C•, ∂) and (D•, ∂) be chain complexes. Show that there is a chain complex
hom• with
homn =

j−i=n
Hom(Ci, Dj)
so that the space of cycles Z0(hom•) = ker(∂ : hom0 → hom−1) is the space
of chain maps C• → D•, and the 0th homology H0(hom•) is the space of chain
maps modulo chain homotopies.
Notation: For abelian groups G1, G2, the set Hom(G1, G2) of group homomor-
phisms G1 → G2 is itself an abelian group under the operation (ϕ, ψ) 7→ ϕ+ ψ
where (ϕ+ ψ)(g) = ϕ(g) + ψ(g).
1
Answered Same Day Mar 05, 2023

Solution

Aditi answered on Mar 05 2023
46 Votes
MATHS SOLUTION
1.
We will use the long exact sequence of the triple (M, B, A), where M is the M¨obius strip obtained from identifying (0, t) with (1, 1 − t) on the square I × I, A is the boundary circle (the image of I × {0, 1} under the quotient map), and B is the image of the boundary of I × I in M.
The long exact sequence of the triple is:
... -> H1(B) -> H1(M) -> H1(M, B) -> H0(A) -> H0(B) -> ...
Since B is a deformation retract of A, we have H1(B) ≅ H1(A) ≅ Z.
We have already shown that H1(M, A) ≅ Z/2Z.
To find H1(M), we note that M deformation retracts onto the circle in the middle of the strip. Therefore, H1(M) ≅ H1(S1) ≅ Z.
Now, consider the map i: B -> M, which is the inclusion map. Since the boundary of the square I × I maps to B, the boundary of the square is homologous to 0 in B. Therefore, the image of the boundary in M is also homologous to 0 in M. This means that the induced map i_*: H1(B) -> H1(M) is the zero map.
So, we have the following portion of the long exact sequence:
... -> H1(M, B) -> H0(A) -> H0(B) -> ...
Since the map i_: H1(B) -> H1(M) is the zero map, the map j_: H1(A) -> H1(M) induced by the inclusion map j: A -> M is an injection. Therefore, H1(M, A) is isomorphic to the quotient group H1(M)/j_*(H1(A)).
To show that the curve f(t) = (0, t) represents the non-zero class in H1(M, A) ≅ Z/2Z, we need to show that its image under the boundary map ∂: H1(M, A) -> H0(A) is non-zero.
Note that the curve f(t) bounds the region of the M¨obius strip on the left of the line x = 0. Let S be this region. We can deform S to the boundary circle A by sliding it along the strip until it reaches the edge, then folding it onto itself along the edge to form a half-twist. This...
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