Given a map of spaces $f:X \to Y$, the mapping cylinder is the adjunction space
$$cyl(f)=(X \times [0,1]) \cup_f Y$$
where we regard $f$ as a map $f: X \times \{1\} \to Y$.\
On the other hand the mapping cone is given by
$$cone(f):=C(X) \cup_f Y$$
where $C(X)=X \times [0,1]/((x,0) \simeq (x',0)$.
Now I'd like to prove that $(cyl(f),X \times 0)$ is a good pair.
How can I wite an explicit homotopy.
I proved that for example $cyl(f) \times 0 \cup X \times 1$ is a deformation retract of $cyl(f) \times [0,1]$. But I do not know if there is a link between these two things.
Furthermore, how could I prove the long exact sequence
$$…H_{n+1}(cone(f)) \to H_n(X) \rightarrow^{f_*} H_n(Y) \to H_n(cone(f)) \to…$$
I also know that the reduced homology of $cone(f)$ is isomorphic to $H_*(cyl(f), X)$.
Best Answer
Regarding your first question, note that for example the neighbourhood $ X \times [0,1/2) \subseteq cyl(f)$ of $X \times {0}$ deformation retracts onto it by the explicit $ (x,t,s) \mapsto (x,(1-s)*t) $.
Regarding the question about the long exact sequence, it follows from the above by the following observations:
The mapping cylinder $ cyl(f) $ deformation retracts onto $ Y $.
The homologies of homotopy equivalent spaces (e.g. deformation retracts) are isomorphic, hence $$ H_n(Y) \simeq H_n(cyl(f)) $$
The good pair discussed above $ (cyl(f), X \times {0}) $ gives rise to a long exact sequence of homology $$ \ldots \rightarrow H_{n+1}(cyl(f),X \times {0})\rightarrow H_n (X \times {0})\rightarrow H_n(cyl(f))\rightarrow H_n(cyl(f),X \times {0}) \rightarrow \ldots $$
Therefore, by using the last comment you wrote down, that is that $$ H_n(cyl(f),X)=H_n(cone(f)) $$ one obtains the long exact sequence above.