Homology isomorphism of $H_n(S^d\times X)$ and $H_{n-1}(S^{d-1}\times X)$

Question

$X$ is an arbitrary space, $d\geq 1$.

The existence of such isomorphism in the title supposedly follows from the Mayer-Vietoris sequence of $(S^d\times X,S^d_{+}\times X,S^d_{-}\times X)$:

$..\rightarrow H_n(S^d_+\times X)\oplus H_n(S^d_-\times X) \rightarrow H_n(S^d\times X)\rightarrow H_{n-1}(S^{d-1}\times X)\rightarrow H_{n-1}(S^d_+\times X)\oplus H_{n-1}(S^d_-\times X)\rightarrow ..$

This is an exact sequence and the homomorphism in the middle should be an isomorphism. But $S^d_+ \times X$ and $S^d_-\times X$ are homotopy equivalent to $X$, so the corresponding homology groups need not be zero. So how can I show this?

Answer 1

Let $X = S^1$ and $d = n = 2$, then $H_2(S^2 \times S^1) = \mathbb{Z}$ but $H_1(S^1 \times S^1) = \mathbb{Z} \oplus \mathbb{Z}$ so you might want to prove something else.