0

I was wondering, may someone please elaborate on why the central core in the mobius band generates the first homology of the mobius band?

My thoughts are: $H_1(M)\cong H_1(\mathbb{S}^1)$ and $H_1(\mathbb{S}^1)$ is generated by the parameterization of the circle $I\rightarrow \mathbb{S}^1$, $t\rightarrow exp(2\pi it)$.

May someone elaborate? How should one think of generators of the first homology group geometrically?

  • 2
    Hint: Find a deformation-retraction of $M$ to the central circle. – Moishe Kohan Nov 17 '21 at 01:41
  • @Moishe Kohan . I am well aware that such a deformation retraction exists. Indeed, that was implicit in the isomorphism between the homology groups above. I don’t understand why that means that a parameterization of the central circle generates thr homology group –  Nov 17 '21 at 01:51
  • 2
    The inclusion map of a deformation retract induces an isomorphism of homology. – Andreas Blass Nov 17 '21 at 01:55

1 Answers1

0

When you say that $H_1(S^1)$ is generated by the parameterization of the circle $u : I\rightarrow S^1, u(t) = \exp(2\pi it)$, you mean that the singular $1$-simplex $u$ is a cycle whose homology class $[u]$ generates $H_1(S^1)$.

The central core (aka center circle) of $M$ is the image of an embedding $j : S^1 \to M$. You know that $S = j(S^1)$ is a strong deformation retract of $M$. Let $r : M \to S$ be the corresponding retraction. This gives us the map $\rho = j^{-1} \circ r : M \to S^1$ which is a homotopy inverse for $j$. Hence $j_*([u])$ is a generator of $H_1(M)$. But $j_*([u]) = [j \circ u]$. The map $u' = j \circ u$ which is given by $u'(t) = j(\exp(2\pi it))$ is a the parameterization of the central circle of $M$. By construction it is a $1$-cycle whose homology class $[u']$ generates $H_1(M)$.

Paul Frost
  • 87,968