Using cellular homology to Comput the homology groups of $S^n \cup_{f} D^{n+1}.$
The question is given below:
Let $f : S^n \rightarrow S^n$ be a map of degree $m.$ Let $ X=S^n \cup_{f} D^{n+1}$ be the space attained from $S^n$ by attaching an $(n+1)$-cell via $f.$ Compute homology groups of $X.$
My questions are:
1- Are we in the case of a good pair?
2- what are the $X^n$ and the $X^{n-1}$ that should be in the LES?
3- will we use lemma 2.34 on pg.137 in AT for calculation?
We use the cellular chain complex $\mathfrak C_*(X)$ of $X$ as described on p.139/140.
We have $\mathfrak C_k(X) = H_k(X^k,X^{k-1})$ which is a free abelian group with basis in one-to-one correspondence with the $k$-cells of $X$, and the boundary maps are given by the Cellular Boundary Formula on p.140.
Let us give $X$ the CW-structure with one $0$-cell, one closed $n$-cell obtained by attaching $D^n$ to the $0$-cell via the constant map on $S^{n-1}$ (thus $X^n = S^n$) and one closed $(n+1)$-cell obtained by attaching $D^{n+1}$ to $S^n$ via a map $f : S^n \to S^n$ of degree $m$.
Thus we have $\mathfrak C_k(X) = \mathbb Z$ for $k = 0,n,n+1$ and $\mathfrak C_k(X) = 0$ else. The cellular boundary formula shows that $d_{n+1}$ is multiplication by $m$. All other $d_k$ are zero. The only case where the latter is is not completely trivial is $k=1$ when $n=1$. This is treated on p.140 before the Cellular Boundary Formula.
Hence $\ker(d_n) = \mathbb Z$ and $\ker(d_k) = 0$ for $k \ne n$. We conclude $$H_0(X) = \mathbb Z$$ $$H_n(X) = \ker (d_n)/\text{im}(d_{n+1}) = \mathbb Z/m\mathbb Z$$ $$H_k(X) = 0 \text{ for } k \ne 0,n$$
Edited on request:
I think the case $m = 0$ is not really in the focus (whence I tacitly assumed $m > 0$). Then $H_{n+1}(X) = \ker(d_{n+1})/\text{im}(d_{n+2}) = \ker(d_{n+1})$. But $d_{n+1}$ is injective, thus its kernel is zero.
Let us have a look at $m = 0$. Then $d_{n+1} = 0$ and $H_{n+1}(X) = \mathbb Z$. Moreover, $H_n(X) = \ker (d_n)/\text{im}(d_{n+1}) = \ker(d_n) = \mathbb Z$. In all other dimensions $k$ it is the same as for $m > 0$.
This is intuitively clear: If the attaching map is constant, then $X = S^n \vee S^{n+1}$.