what is the homology groups of a contracting circle in a torus to a point?

I need to compute the homology groups of the space $$X$$ obtained by contracting a circle in a torus $$T$$ to a point.

I think $$H_0(X)= \mathbb{Z}$$ since $$X$$ is a connected. For $$H_1(X)$$ I think we need to calculate the fundamental group which is $$\mathbb{Z}\times \mathbb{Z}$$, and it is $$H_1(X)=\mathbb{Z}\times \mathbb{Z}$$. For $$H_2(X)$$, I believe it is $$\mathbb{Z}$$.

I really need help here, any help or comment would be appreciated.

• Did you have any particular circle in mind? I ask because the outcome is likely to depend on what circle you collapse. Apr 9, 2022 at 22:25
• @LeeMosher, not really. Apr 9, 2022 at 22:34

Let $$C$$ be some essential embedding of $$S^1$$ in the torus. Write it as $$(m,q)$$ for the image of $$1 \in \mathbb Z$$ under the induced map $$H_1(S^1) \to H_1(T)$$ where $$m$$ is coprime to $$q$$. Here are some fun ways to see the result.

1. You can use the idenfitication $$\tilde{H_n}(X)=H_n(T,C)$$ and the short exact sequence to see that we have

$$0 \to H_2(T) \to \tilde{H}_2(X) \to H_1(S^1) \to H_1(T) \to \tilde{H}_1(X) \to H_0(S^1) \to H_0(T) \to 0$$

Where the map $$H_1(S^1)= \mathbb Z \to \mathbb Z \oplus \mathbb Z= H_1(T)$$ is given by $$n \mapsto (nm,nq)$$.

1. You can use the classification of surfaces with boundary to deduce that $$T \setminus C \cong S^1 \times I$$, and from Alexander-Lefschetz duality we know that $$\tilde{H}_{n-k}(X)=H_{n-k}(T,C) \cong H^{k}(S^1)$$.

2. There is a homeomorphism of the torus carrying $$C$$ to $$(1,0)$$ in the torus (basically because $$T \setminus X$$ is homeomorphic to $$S^1 \times I$$ you can use the homeomorphism there, and glue back together.) You can then argue geometrically that $$X \simeq S^2 \vee S^1$$ and use the Mayer-Vietoris sequence to compute its homology. A hint for this homotopy equivalence is that $$X$$ is homotopy equivalent to a sphere with two points identified.

If the curve $$C$$ is not essential, i.e. bounds a disk, we can see that $$X=T \vee S^2$$ and compute the homology this way, or use the long exact seqeuence from $$1$$, except that $$H_1(S^1) \to H_1(T)$$ will be the zero map

• Probably worth mentioning that in this exercise, its probably intended that $C=(1,0)$ or $(0,1)$. You can then use $1$ or $3$ in a way where the argument is a bit simpler. Apr 10, 2022 at 3:11
• There is another case when the circle bounds a disk in the torus. Apr 10, 2022 at 7:14
• @CheerfulParsnip oh shoot :(. I’ll edit in a bit haha Apr 10, 2022 at 16:46