Suppose $f: S^{2n-1}\to S^n$ a continuous and $X = S^{n} \bigcup\limits_{f}e^{2n},$ where $e^{2n}$ is $2n$-cell homeomorphic to the open $2n$-disk int$(D^{2n}).$
My question is why $H_k(X) = \left\{\begin{array}{ll} \mathbb{Z}&\mbox{if }k = 0, n,2n\\ 0&\mbox{otherwise}. \end{array}\right.$
Can you explain for me? Thank you very much!
If you look at the chain complex that calculates cellular homology, it has $0$ in all dimensions, except for $k = 0, n, 2n$, where it has a free group $\cong \mathbb{Z}$ generated by the corresponding cell. Such complex clearly has homology as above.
$$0 \to \mathop{\mathbb{Z}}_{2n} \to 0 \to \cdots \to 0 \to \mathop{\mathbb{Z}}_n \to 0 \to \cdots \to 0 \to \mathop{\mathbb{Z}}_0 \to 0$$
I assume $n > 1$. For $n = 1$ the claim is false: glue $D^2$ to the circle to obtain the projective plane $\mathbb{RP}^2$. The corresponding chain complex is $$0 \to \mathop{\mathbb{Z}}_2 \xrightarrow{\times 2} \mathop{\mathbb{Z}}_1 \xrightarrow{0} \mathop{\mathbb{Z}}_0 \to 0$$