Definition $:$ Let $X,Y$ be topological spaces and $f : X \longrightarrow Y$ be a map. The mapping cylinder $M_f$ of $f$ is obtained from the disjoint union $(X \times I) \sqcup Y$ by identifying $(x,1) \sim f(x),$ for all $x \in X.$
Exercise $:$ Show that $M_f$ deformation retracts to $Y.$
I think if we can take every point of $M_f$ and squeeze it down to it's image under $f$ via straight line deformation I would get a deformation retract of $M_f$ onto $f(X).$ But I am not sure as to why $M_f$ deformation retracts to $Y.$ Would anybody provide me an explicit formula for such deformation? I am not satisfied with hand-waving description of algebraic topology; rather I want to understand it analytically. That's why I am asking for analytic formula for such deformation. Any help in this regard will be appreciated.
Thanks in advance.
Best Answer
Quite simply, notice that there is a straight line connecting $(x,0)$ and $(x,1)$ given by $f_x(t)=(x,t).$ In fact, $f_x$ is continuous as for both projection maps $\pi_X$ and $\pi_I,$ we have that $\pi_X\circ f_x$ and $\pi_I\circ f_x$ are continuous. Now, consider the map $H:X\times I\times I \rightarrow Y$ given as,
$$H((x,s),t)=(x,s(1-t)+t).$$
Notice that $H$ is a deformation retraction onto $Y.$