# mapping cone and cylinder

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 $H_*(cyl(f), X)$.

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 $(x,t,s) \mapsto (x,(1-s)*t)$.
1. The mapping cylinder $cyl(f)$ deformation retracts onto $Y$.
2. The homologies of homotopy equivalent spaces (e.g. deformation retracts) are isomorphic, hence $$H_n(Y) \simeq H_n(cyl(f))$$
3. 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.