In Hatcher example 3B.3, we calculate the homology of $X\times S^k$. I know there is a solution by Kunneth formula $$ H_n(X\times S^k)\cong \bigoplus\limits_{n=r+s}H_r(X)\otimes_\mathbb Z H_s(S^k) $$ where this boils down to $H_n(X)\oplus H_{n-k}(X)$ by property of the tensor product and homology. Hatcher seems to calculate the homology by looking at the chain complex, as follows:
The boundary formula in $C_*(X\times S^k)$ takes the form $d(a\times b) = da\times b$ since $d = 0$ in $C_*(S^k)$. Therefore, the chain complex $C_*(X\times S^k)$ is just the direct sum of two copies of the chain complex $C_*(X)$, one of the copies having its dimension shifted upwards by $k$. Therefore, $H_n(X\times S^k)\cong H_n(X)\oplus H_{n-k}(X)$.
Can anyone explain the italic part, where we obtain a direct sum of the chain complexes with some index shifting? It is unclear to me how this part was done. Thank you!
Hatcher works with the cellular homology groups of CW-complexes. In Proposition 3B.1. he proves that the boundary map in the cellular chain complex $C_*(X×Y)$ is determined by the boundary maps in the cellular chain complexes $C_∗(X)$ and $C_∗(Y)$ via the formula $d(e_i×e_j) = de_i×e_j + (−1)^ie_i×de_j$.
Now consider $Y = S^k$ with the standard CW-structure (one $0$-cell $b^0$ and one $k$-cell $b^k$). Then $d = 0$ in $C_*(S^k)$. This is trivial for $k \ne 1$. For $k = 1$ we enconter a little problem because the cellular boundary formula (see p.140) requires to determine the degree of a map $S^0 \to S^0$. For the answer see In cellular homology, how should we define the degree of a map between $0$-spheres? It shows that $d=0$ also in this case.
Therefore $d(a \times b) = da \times b$ for the cells of $X \times S^k$.
The $n$-cells of $X \times S^k$ are the products $a \times b$ of all cells $a$ of $X$ and $b$ of $S^k$ such that $\dim a + \dim b = n$. Therefore the basis of the free abelian group $C_n(X \times S^k)$ is $$\mathcal B_n(X \times S^k) = \mathcal B_n(X) \times \{b^0\} \cup \mathcal B_{n-k}(X) \times \{b^k\} $$ where $\mathcal B_i(X)$ denotes the set of $i$-cells of $X$. Using the above boundary formula this shows that $$C_n(X \times S^k) \approx C_n(X) \oplus C_{n-k}(X).$$ Since