# Short exact sequence of abelian groups how to cancel $\mathbb Z$ in the middle term.

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.

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.