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?
2026-03-27 04:38:44.1774586324
The effect of attaching the Möbius strip to the torus
666 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
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.