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

$X$ 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?

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Since $H_n(S^d\times X)\cong H_n(X)\oplus H_{n-d}(X)$, the statement is generally false. – Carsten S Feb 21 '14 at 11:39
Thanks. Maybe I made a mistake copying the statement. – user35359 Feb 21 '14 at 11:45
One way to fix it is to consider the kernel of the map $H_n(S^d\times X)\to H_n(X)$ induced by the projection map instead. – Carsten S Feb 21 '14 at 12:07
The spaces $S^d_+\times X$ and $X$ are homotopy equivalent, thus the map you are considering is an isomorphism in reduced homology if, and only if $X$ is acyclic. – Daniel Robert-Nicoud Feb 22 '14 at 0:36

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.