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

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...