# Finding a mistake using Mayer-Vietoris

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?

• How exactly is $\coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points? Feb 28, 2019 at 14:42
• @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 $n=3,k=1$.
– Javi
Feb 28, 2019 at 14:44
• @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.
– Javi
Feb 28, 2019 at 14:47

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 $$S^2$$.