# Relative Homology of the Mapping Cylinder w.r.t a Subspace

Given a continuous map $f:X\to Y$, the mapping cylinder of $f$ is defined as the space obtained from $(X\times I)\sqcup Y$ by identifying $(x, 1)$ with $f(x)$ for all $x$.

Let $f:S^n\to S^n$ be a degree $m$ map. Let the domain sphere we written $s^n_d$ and the target sphere be written $S^n_t$.

On pg. 148 of Hacther's Algebraic Topology, the author writes that $H_n(M_f, S^n_d)\cong \mathbf Z/m\mathbf Z$ but does not give a proof.

I am unable to prove this. I thought of using the fact that the mapping cylinder deformation retracts to $S^n_t$. But this only gives us that $H_n(M_f)\cong \mathbf Z$.

• Hmm, how about using the long exact sequence of the pair?
– Hmm.
Jun 18 '16 at 9:23

As suggested by @SoumyaSinha, suppose $n\ge2$ and consider the long exact sequence of $(M_f,S^n_d)$, which around degree $n$ is: $$0\longrightarrow H_n(S_d^n)\longrightarrow H_n(M_f)\longrightarrow H_n(M_f,S_d^n)\longrightarrow0.$$ under the deformation retract $M_f\to S_t^n$, the first map simply becomes $f$, which we know to have degree $m$, and therefore the sequence becomes $$0\longrightarrow \mathbb{Z}\stackrel{m}\longrightarrow \mathbb{Z}\longrightarrow H_n(M_f,S_d^n)\longrightarrow0$$ and we conclude $H_n(M_f,S_d^n)\cong\mathbb{Z}/m\mathbb{Z}$.