Suppose I am given an exact sequence:

$$0\to G\xrightarrow{f} \mathbb{Z} \xrightarrow{g} \mathbb{Z} \xrightarrow{h} H\to 0 $$

where the first $\mathbb{Z}=H_3(A\cup B)$ and the second $\mathbb{Z}=H_2(A\cap B)$ where $A,B$ are solid tori. I compute the homology of $A\cup B$ and $A \cap B$ by noting that $A\cup B=S^3, A\cap B=$ Torus.

I know by exactness that $f$ should be injective, i.e. $ker(f)=0$ and $h$ should be surjective, i.e. $im(h)=H$. But I got stuck when computing $G$ and $H$.

Any help is appreciated :)

  • $\begingroup$ Where did $G$ and $H$ and the two zeroes on the ends come from? Did this all come from $$H_3(A \cap B) \to H_3(A) \oplus H_3(B) \to H_3(A \cup B) \to H_2(A \cap B) \to H_2(A) \oplus H_2(B) \to H_2(A \cup B)$$? If so, you have not made full use of the Mayer-Vietoris sequence. Since $A,B$ are solid tori, they are homotopy equivalent to circles, and therefore $H_2(A)=H_3(A)=H_2(B)=H_3(B)=0$. $\endgroup$
    – Lee Mosher
    May 24, 2016 at 14:08
  • $\begingroup$ @LeeMosher. Yap, that is the full version of Mayer Vietoris sequence. I know the fact that solid torus is homotopic to 1-sphere hence the homology groups are the same. But I am supposed to use the MVS to compute that. $\endgroup$ May 24, 2016 at 14:24

1 Answer 1


From a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$ which is easier than the case you describe, knowing $A$ and $B$ in general is not enough to compute $C$. This is the extension problem.

Some examples $$ 0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0\rightarrow 0, $$

or $$ 0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}\rightarrow 0.$$

You should know some more information of the maps in the sequence to go further with your computation.


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