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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.

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

1 Answer 1

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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

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  • $\begingroup$ 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. $\endgroup$ Apr 10, 2022 at 3:11
  • $\begingroup$ There is another case when the circle bounds a disk in the torus. $\endgroup$ Apr 10, 2022 at 7:14
  • $\begingroup$ @CheerfulParsnip oh shoot :(. I’ll edit in a bit haha $\endgroup$ Apr 10, 2022 at 16:46

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