I was computing the homology of $S^3-\coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $\mathbb{R}^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $\mathbb{R}^2$ and finaly the last one ends up with a space homotopic to $S^1$. Therefore, $H_2(S^3-\coprod_{i=1}^4 I_i)=0$.

But in other calculations this caused me some problems so I decide to do it formally using Mayer-Vietoris. I had already computed $H_*(S^3-I\sqcup I)$, giving me a consistent result with the intuition above, i.e. $H_2(S^3-I\sqcup I)=\mathbb{Z}$.

Now I decompose $S^3=(S^3-\coprod_{i=1}^2 I_i)\cup (S^3-\coprod_{i=3}^4 I_i)$. From Mayer-Vietoris there is a short exact sequence

$$0\to H_3(S^3)\to H_2(S^3-\coprod_{i=1}^4 I_i)\to H_2(S^3-I_1\sqcup I_2)\oplus H_2(S^3-I_3\sqcup I_4)\to 0$$

From my calculations this would be

$0\to\mathbb{Z}\to H_2(S^3-\coprod_{i=1}^4 I_i)\to \mathbb{Z}\oplus\mathbb{Z}\to 0$

But then $H_2(S^3-\coprod_{i=1}^4 I_i)\neq 0$, which is inconsistent with my first reasoning. Where is the mistake?

  • 2
    $\begingroup$ How exactly is $\coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points? $\endgroup$ Feb 28, 2019 at 14:42
  • $\begingroup$ @Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $\mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $\widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$. $\endgroup$
    – Javi
    Feb 28, 2019 at 14:44
  • $\begingroup$ @Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere. $\endgroup$
    – Javi
    Feb 28, 2019 at 14:47

1 Answer 1


By your argument at the beginning, removing four points from a two-sphere should yield something which is homotopy equivalent to a zero-sphere. This is obviously not true. Your error is that after getting a space homotopy equivalent to $S^2$, you proceed as if it is actually homeomorphic to $S^2$.

  • 1
    $\begingroup$ Oh that's it, I was so dumb! Thank you $\endgroup$
    – Javi
    Feb 28, 2019 at 15:00
  • 12
    $\begingroup$ @Javi, happens to all of us. $\endgroup$
    – Carsten S
    Feb 28, 2019 at 15:01

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