Now I want to calculate homology group of some topological space.

Using Mayer-Vietoris sequence I end up with the following short exact sequence:

$k$ :a knot

$$0\to H_0(Torus)\to H_0(\mathbb R^3\setminus k)\oplus H_0(S^1)\to H_0(\mathbb R^3)\to 0$$

Following maps are irrelevant to my question but for completeness:

First map is $(j_U,-j_V):\mathbb Z\to H_0(\mathbb R^3\setminus k) \oplus \mathbb Z$

Second map is $g_U\oplus g_V: H_0(\mathbb R^3\setminus k) \oplus \mathbb Z \to \mathbb Z$

Where $j_U,j_V,g_U,g_V$ are canonical inclusion maps.

Since in above short exact sequence $H_0(Torus)=H_0(S^1)=\mathbb Z$ free abelian groups then $$H_0(\mathbb R^3\setminus k) \oplus \mathbb Z=\mathbb Z\oplus \mathbb Z$$

I know that $H_0(\mathbb R^3\setminus k)$ is $\mathbb Z$ since it is path connected but Iam asking this question to understand the real algebraic reason:

How can we cancel $\mathbb Z$ and say $H_0(\mathbb R^3\setminus k)=\mathbb Z$

  • Should I use explicitly the canonical maps
  • Is there any usefull theorem that if I encounter with some sequence like that and I want to cancel some group like this, it helps me.

1 Answer 1


Finitely generated abelian groups are cancellable among abelian groups, i.e. if $A$ is a finitely generated abelian group and $B,C$ are abelian groups such that $A\oplus B\cong A\oplus C$, then $B\cong C$. This is proven e.g. in this paper. However, I would advise against relying on this fact, as there's much concrete and canonical reasons for this conclusion to hold in the present context. Here are three options:

  1. Use the Mayer-Vietoris sequence for reduced instead of unreduced homology. This immediately yields $\tilde{H}_0(\mathbb{R}^3\setminus k)=0$, whence $H_0(\mathbb{R}^3/k)=\mathbb{Z}$ by the general relationship between reduced and unreduced homology.

  2. Recall that if $X$ is any topological space and $\pi_0(X)$ denotes the set of path-components of $X$, there is a canonical isomorphism $H_0(X)\cong\mathbb{Z}^{(\pi_0(X))}$ (the latter meaning the free abelian group with generators $\pi_0(X)$). Using these identifications, your exact sequence becomes $$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}^{(\pi_0(\mathbb{R}^3\setminus k))}\oplus\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow0.$$ The latter map is addition, since the inclusion takes any path-component $\mathbb{R}^3\setminus k$ or $S^1$ to the unique path-component of $\mathbb{R}^3$. Thus, its kernel consists of those elements whose entries add up to $0$. By exactness, this kernel is the image of the map $\mathbb{Z}\rightarrow\mathbb{Z}^{(\pi_0(\mathbb{R}^3\setminus k))}\oplus\mathbb{Z}$, but the image of this map meets only one summand of $\mathbb{Z}^{(\pi_0(\mathbb{R}^3\setminus k))}$, because the torus is path-connected. Together, these conditions force that $|\pi_0(\mathbb{R}\setminus k)|=1$, so $H_0(\mathbb{R}^3/k)=\mathbb{Z}$ is free on the unique path-component of $\mathbb{R}^3\setminus k$.

  3. Note that you can split your short exact sequence by the map $H_0(\mathbb{R}^3)\rightarrow0\oplus H_0(S^1)\subseteq H_0(\mathbb{R}^3\setminus k)\oplus H_0(S^1)$, mapping the canonical generator of $H_0(\mathbb{R}^3)$ to the canonical generator of $H_0(S^1)$. This induces an isomorphism $H_0(T)\oplus H_0(\mathbb{R}^3)\cong H_0(\mathbb{R}^3\setminus k)\oplus H_0(S^1)$ with the additional property that $0\oplus H_0(\mathbb{R}^3)$ corresponds to $0\oplus H_0(S^1)$. This trivializes the cancellation problem (in contrast to the general situation addressed at the start), as we can calculate $$H_0(\mathbb{R}^3\setminus k)\cong\frac{H_0(\mathbb{R}^3\setminus k)\oplus H_0(S^1)}{H_0(S^1)}\cong\frac{H_0(T)\oplus H_0(\mathbb{R}^3)}{H_0(\mathbb{R}^3)}\cong H_0(T)\cong\mathbb{Z}.$$ Here, the middle isomorphism is induced by the splitting we constructed.


You must log in to answer this question.

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