Homology group of mapping cylinder for map of degree m

Question

Hatcher claims on p. 148 that $H_{n}(M_{f},S^{n})=\mathbb{Z}/m\mathbb{Z}$, where $f$ is a degree $m>1$ map of $S^{n}$, and $M_{f}$ is the associated mapping cylinder. Why is this?

Answer 1

Firstly, you have to be more specific about how you intended $S^n$ to embed in $M_f$. I didn't look in Hatcher but I'm supposing it's the $S^n$ on the "0" end of the cylinder. If it's the $S^n$ on the "1" end, then $H_n(M_f, S^n)$ is trivial.

Think about the long exact sequence in which relative homology fits:

$$ \cdots H_{n}(S^n) \stackrel{\alpha}{\longrightarrow} H_{n}(M_f) \stackrel{\beta}{\longrightarrow} H_{n}(M_f, S^n) \stackrel{\gamma}{\longrightarrow} H_{n-1}(S^n) = 0 $$

One can see that $\beta$ is a surjection, so $$ H_n(M_f, S^n) = H_n(M_f) / \mathrm{im} (\alpha). $$ Can you see how this proves your result? Think about what $\alpha$ does to a generator of $H_n(S_n)$.