Let $U$ be an open set which is constructed as intersection of $S^1$ and open ball in $\mathbb{R}^2$. And $x$ is just a point contained in $U$.

My opinion: By long exact sequence, $H_n(U, U-x)$ is zero. Because $H_n(U)$ is zero since $U$ is contractible and $H_n(U-x)$ is contractible, too.

Why is this contractible? Actually $U-x$ is same with union of $U_1$ and $U_2$, which are also open set of $S^1$ and homeomorphic to $U$. So $H_n(U-x)$ is isomorphic to direct sum of $H_n(U_1)$ and $H_n(U_2)$ and they are zero, so $H_n(U-x)$ is zero, too.

But real answer is $H_n(U, U-x)$ is isomorphic to $\mathbb Z$; set of integers. Why my opinion is not true? Can anyone find my errors?


You are forgetting about $H_0$. Your long exact sequence is $$ \cdots\to H_n(U-x)\to H_n(U)\to H_n(U,U-x)\to \cdots\to H_0(U-x)\to H_0(U)\to H_0(U,U-x)\to 0 $$ Now $H_0(U-x)\simeq\mathbb{Z}\oplus \mathbb{Z}$ and $H_0(U)\simeq\mathbb{Z}$. If you do this explicitly you will see that kernel of $H_0(U-x)\to H_0(U)$ is isomorphic to $\mathbb{Z}$. On the other hand, since $H_1(U)=H_1(U-x)=0$ you have an exact sequence $$ 0\to H_1(U,U-x)\to H_0(U-x)\to H_0(U) $$ now since kernel is also $\mathbb{Z}$, you find $H_1(U,U-x)\simeq \mathbb{Z}$. For $n>1$, since $H_{n-1}(U-x)=H_n(U)=0$, the exact sequence reduces to $0\to H_n(U,U-x)\to 0$ meaning $H_n(U,U-x)=0$ for $n>1$.

Edit: I made a mistake. The cokernel of $H_0(U-x)\to H_0(U)$ is in fact $0$, see Hatcher exercise 2.1.16 (a): "$H_0(X,A)=0$ iff $A$ meets every path component of $X$". This show that $H_0(U, U-x)=0$ too. The only non-zero homology group is $H_1(U, U-x)\simeq \mathbb{Z}$.

  • $\begingroup$ $H_n(U,U-x)$ is actually $0$ for $n>1$. $\endgroup$ Jun 23 '16 at 16:57
  • $\begingroup$ Yeah I just realized that myself... $\endgroup$
    – Hamed
    Jun 23 '16 at 16:59
  • $\begingroup$ Hmm, maybe I was making mistake. you're correct $\endgroup$
    – Wylnorr
    Jun 23 '16 at 17:08
  • $\begingroup$ I found my opinion is actually correct, just read something wrong... $\endgroup$
    – Wylnorr
    Jun 23 '16 at 17:10
  • $\begingroup$ Of course your opinion is absolutely correct, other than the $H_0$ part. I haven't done anything in my answer but your opinion actually. $\endgroup$
    – Hamed
    Jun 23 '16 at 17:17

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