Obtaining an isomorphism from a relative Mayer-Vietoris sequence

Question

This is another question from an old qual exam. Show with the relative Mayer-Vietoris sequence that if $X$ is a finite polyhedra, $p_0 \in S^r$, and $k >r$, then $H_k(X \times S^r, X \times \{p_0\}) \cong H_{k-r}(X)$. For the relative Mayer-Vietoris sequence for $(X_1 \cup X_2, A_1 \cup A_2)$, where $A_i \subset X_i$, I've tried a couple of different options for the $X_i$ and $A_i$, but I can't see how to get a dimension-lowering isomorphism from this sequence.

Answer 1

Without loss of generality, suppose $p_0$ lies in the equator of $S^r$. Write $S^r = D_{+}^r \cup D_{-}^r$ where $D_{+}^r =\{(x,t)\in S^r : t > -1/2\}$ and $D_{-}^r = \{(x,t)\in S^r : t < 1/2\}$. The intersection $D_{+}^r \cap D_{-}^r$ deformation retracts to the equator $S^{r-1}$. In the relative Mayer-Vietoris sequence take $X_1 = X \times D_{+}^r$, $X_2 = X \times D_{-}^r$, and $A_1 = A_2 = X\times \{p_0\}$; the sequence contains the exact sequence

\begin{align} &H_k(X\times D_{+}^r, X\times \{p_0\}) \oplus H_k(X\times D_{-}^r, X \times \{p_0\})\rightarrow H_k(X\times S^r, X\times \{p_0\})\\ &\xrightarrow{\partial} H_{k-1}(X\times S^{r-1}, X\times \{p_0\}) \to H_{k-1}(X\times D_{+}^r, X\times \{p_0\}) \oplus H_{k-1}(X\times D_{-}^r, X \times \{p_0\})\end{align}

Since $D_{+}^r$ and $D_{-}^r$ deformation retract to $\{p_0\}$, the first and last groups in the above sequence are trivial, making the boundary map an isomorphism: $$H_k(X\times S^r, X\times \{p_0\})\xrightarrow[\cong]{\partial} H_{k-1}(X\times S^{r-1}, X\times \{p_0\})$$ Now work inductively to obtain the result.