Let $f: S^1 \times D^2 \to S^3 $ be an embedding, and let $g=f|_{S^1 \times 0}$ denote the induced embedding of a circle, and let $K=g(S^1)$. I want to compute the homology groups of $A:=S^3-K$ using the Mayer-Vietoris sequence.

Since $A$ is path-connected (by, for example, the Jordan curve theorem), $H_0(A)=\Bbb Z$.

Let $B=f(S^1 \times D^2)$. Since $f$ is an embedding, $B$ is homeomorphic to a solid torus, hence homotopy equivalent to a circle, so we know the homology groups of $B$. Also, we have $A \cup B =S^3$. Now we consider the Mayer-Vietoris sequence for the decomposition $S^3 = A \cup B$ (Actually, we have to consider an open neighborhoods of $B$ that deformation retracts onto $B$, instead of $B$). Note that $A \cap B$ is homotopy equivalent to a torus. Then it is easily shown that $H_k (A)=0$ for all $k >3$. So we only need to consider $k=1,2,3$.

For $k=1$, it is quite easy. We have:

$H_2(S^3)=0 \to H_1(T^2)= \Bbb Z^2 \to H_1(A) \oplus H_1(S^1)=H_1(A) \oplus \Bbb Z \to H_1(S^3)=0.$

Thus, $H_1(A)=\Bbb Z$.

For $k=2,3$ we need to consider the following:

$H_3(T^2)=0 \to H_3(A) \oplus H_3(S^1)=H_3(A) \to H_3(S^3)=\Bbb Z \to H_2(T^2)=\Bbb Z \to H_2(A)\oplus H_2(S^1)=H_2(A) \to H_2(S^3)=0$.

If we can examine the map $H_3(S^3) \to H_2(T^2)$, we will be done, but I have no idea here. How do I have to proceed?

On the other hand, is there another method, other than the Mayer-Vietoris, to compute the homology groups of $A$?

  • 1
    $\begingroup$ Alexander duality? $\endgroup$ Commented Dec 14, 2019 at 17:40
  • $\begingroup$ @LordSharktheUnknown I don't know about it $\endgroup$
    – blancket
    Commented Dec 14, 2019 at 17:40
  • $\begingroup$ en.wikipedia.org/wiki/Alexander_duality $\endgroup$ Commented Dec 14, 2019 at 17:45
  • $\begingroup$ @LordSharktheUnknown Wow That makes the question trivial $\endgroup$
    – blancket
    Commented Dec 14, 2019 at 17:49
  • 1
    $\begingroup$ Also see Dold, Albrecht. "A Simple Proof of the Jordan-Alexander Complement Theorem." The American Mathematical Monthly 100, no. 9 (1993): 856-57. doi:10.2307/2324661. $\endgroup$ Commented Dec 14, 2019 at 17:51

2 Answers 2


While it is indeed very quick to do this with Alexander duality, you can still do this with the Mayer-Vietoris sequence, and with a tiny bit of Poincare duality.

As you say, the problem is quickly finished if one can evaluate the connecting homomorphism
$$\underbrace{H_3(S^3)}_{\mathbb Z} \xrightarrow{\delta} \underbrace{H_2(T^2)}_{\mathbb Z} $$ This map $\delta$ is an isomorphism (from which it follows quickly that $H_2(A) \approx 0$).

To deduce this, you need to use the definition of $\delta$ in the setup of the Mayer-Vietoris sequence, and you need to know how the generators of $H_3(S^3)$ and $H_2(T^2)$ are defined in the early parts of the proof of Poincare duality.

Triangulate $S^3$ so that $A$, $B$ and $T^2 = A \cap B$ are subcomplexes.

Poincare duality tells you that the generator of $H_3(S^3)$ is a homology class $[c]$ represented by a 3-cycle $c$ that assigns constant coefficient $+1$ to each 3-simplex of $S^3$ (assuming that those 3-simplices are assigned orientations compatible with a global orientation of $S^3$).

To define $\delta[c] \in H_2(T^2)$, here's what you do. Restrict $c$ to either $A$ or to $B$ (it doesn't matter which, up to sign), let's say to $A$, and denote the restricted 3-cycle as $c_A$. Its boundary $\partial c_A$ is a 2-cycle supported on $T^2$. By definition $$\delta[c] = [\partial c_A] $$ From the construction it's pretty clear that $\partial c_A$ assigns constant coefficient $+1$ (or $-1$) to each 2-simplex on $T^2$. Hence, again by Poincare duality, $[\partial c_A]$ is a generator of $H_2(T^2) \approx \mathbb Z$.

  • $\begingroup$ Thanks for the correction, it's nice to have old stupid errors fixed! $\endgroup$
    – Lee Mosher
    Commented May 16, 2021 at 14:34

As another method that uses only Mayer-Vietoris, we can use different subspaces.

Split $K$ into a pair of arcs $a, b \subset S^3$ that meet at a pair of points $p_1,p_2$. Then the open subspaces $A = S^3 \setminus a$ and $B = S^3 \setminus b$ cover the space $X = S^3 \setminus \{p_1,p_2\}$. In this case, the space you are interested in is $S^3 \setminus K = A \cap B$. Both $A$ and $B$ are contractible, and $X$ is a homotopy $2$-sphere, so the Mayer-Vietoris sequence gives

$$\widetilde H_{n+1}(S^2) \cong \widetilde H_n(S^3 \setminus K)$$

In other words, $S^3 \setminus K$ is a homology $1$-sphere.


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