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Let $A$ be a point on a circle in $\mathbb{R}^2$ and let $B$ be the complement of $A$ in this circle. How do I show that $\{A,B\}$ is not an excisive pair?

I say wlog, let $U=\{z\in \mathbb{R}^2:|z|=1\}$, $A=\{(1,0)\}$, and $B=U-A$. Then we claim that $h:S_*(A;\mathbb{Z})\oplus S_*(B;\mathbb{Z})\to S_*(U;\mathbb{Z})$ defined by $h(ra\oplus r'b)=ra+r'b$ is not a chain homotopy equivalence.This means that there exists an $n\in\mathbb{Z}$ such that $H_n(h):H_n(S_*(A;\mathbb{Z})\oplus S_*(B;\mathbb{Z}))\to H_n(S_*(U;\mathbb{Z}))$ is not an isomorphism.

I am struggling because I have no idea how to compute $H_n(S_*(A;\mathbb{Z})\oplus S_*(B;\mathbb{Z}))$.I know that $H_n(S_*(A;\mathbb{Z})\oplus S_*(B;\mathbb{Z}))\cong H_n(S_*(A;\mathbb{Z})\oplus H_n(S_*(B;\mathbb{Z})$ and computing $H_n(S_*(A;\mathbb{Z})$ is easy. I have no idea how to compute $H_n(S_*(B;\mathbb{Z})$.

Johnathon Taylor
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1 Answers1

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Both $A$ and $B$ are nonempty and contractible. Therefore $$H_0(A)\cong H_0(B)\cong\Bbb Z$$ and $$H_n(A)\cong H_n(B)=0$$ for $n\ge1$. If they were an excisive pair, then the Mayer-Vietoris sequence would be exact.

Part of this is $$H_1(A)\oplus H_1(B)\to H_1(A\cup B)\to H_0(A\cap B).$$ But that is $$0\to\Bbb Z\to 0$$ as $A\cup B$ is a circle whose $H_1$ is $\Bbb Z$.

Angina Seng
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