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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you very much that really solves my confusion! $\endgroup$
    – Alex
    Commented Mar 13, 2019 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .