I don't understand the following fact I've read:

Any map $f:X \rightarrow Y$ can be written as composition $X \stackrel{i}{\hookrightarrow} M_f \stackrel{j}{\rightarrow} Y$, where $i$ is the inclusion as a subset,$j$ is an homotopy equivalence and $M_f$ is the mapping cylinder of $f$. How could be checked this?

Thank you in advance!


1 Answer 1


The mapping cylinder will be defined as something like $M_f=X\times[0,1]\coprod Y/\sim$, where $\sim$ is defined by $(x,1)\sim f(x)$. Including $X$ via the map $i:X\hookrightarrow M_f$ sinding $x$ to $(x,0)$ is an obvious inclusion and the harder part is the homotopy equivalence.

What we can, in fact, show, is a fair bit stronger: $M_f$ strongly deformation retracts onto $Y$, that is, there is a map $F:M_f\times I\to M_f$ such that $F(y,t)=y$ for all $y\in Y$, $F((x,t),0)=(x,t)$ and $F((x,t),1)\in Y$ for all $(x,t)\in M_f$. One can find such a map by $F:((x,t),s)\mapsto (x,t+(1-t)s)$ for $(x,t)\in X\times I$ (and $F|Y=\text{id}_Y$). We can now use this to define the map $j$, namely $j=F|M_f\times \{1\}$.

To see that this is a homotopy equivalence, we have to use the inclusion $\iota:Y\to M_f$ and show that $\iota\circ j$ and $j\circ\iota$ are homotopic to their respective identities. One of these simply equals $\text{id}_Y$, so there's not much work there and for the other one, the map $F$ that we just carefully constructed can be used. (In fact, this proves that a (strong) deformation retract is a homotopy equivalence.)


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