I am trying to show the following:
If $f : X \to Y$ is a homotopy equivalence and moreover a cellular map between finite CW-complexes, then $X$ is a deformation retract of $M_f$ via a cellular map. Here $M_f$ denotes the mapping cylinder of $f$.
To this end, we need to show that there exists a map $F : M_f \times I \to M_f$ such that $F(m,0)=m$, $F(x,1)=x$ and $F(m,1) \in X$, where $m \in M_f$ and $x \in X \subset M_f$ (for simplicity we identify $X \cong X \times \{1\} \subset M_f$).
So far, I have the following: we know that $f$ factors as
$$
f = q \circ j : X \hookrightarrow M_f \to Y,
$$
where $j$ is the canonical inclusion and $q$ is a homotopy equivalence, so in particular we get: there exists a map $p : Y \to M_f$ and a homotopy $H : M_f \times I \to M_f$ with $H_0 = \text{id}_{M_f}$ and $H_1 = p \circ q$.
Then I thought I could simply define $F(m,t) := H_t(m)$, since then it would already follow that
$$
F(m,0)=H_0(m)=\text{id}_{M_f}(m)=m \quad\text{and}\quad F(x,1)=H_1(x)=p(q(x))=p(f(x))=j(x)=x,
$$
where we used that $f=q\circ j$. However, for the last calculation we get $F(m,1) = H_1(m) = p(q(m))$, and we'd like to show that this lies in $X$, but I do not see how to do this.
My questions:
- Defining $F$ this way seemed promising. Am I on the right track?
- I am able to show that $j : X \hookrightarrow M_f$ is a cofibration. I read that because $f$ and $q$ are homotopy equivalences it follows that $j$ is a homotopy equivalence as well. Can we use this to show that $j$ is a deformation retract?
- I have not been able to include the fact that $f$ is a cellular map. How can I fix this?
Any hints on where I'm going wrong or how to save the proof are warmly appreciated!
Best Answer
Your approach does not work. Indeed you must use fact that $f$ is a homotopy equivalence.
Note that we identify $X$ with the set $\{[x,1] \mid x \in X \} \subset M_f$ and $Y$ with the set $\{[y] \mid y \in Y \} \subset M_f$.
Under your assumptions the cellular approximation theorem implies that there exists a cellular $g : Y \to X$ such that $g \circ f \simeq id_X$ and $f \circ g \simeq id_Y$ via cellular homotopies.
$X$ is a retract of $M_f$ if $f$ has a left homotopy inverse (this is a map $g : Y \to X$ such that $g \circ f \simeq id_X$).
Let $L : X \times I \to X$ be a homotopy such that $L(x,0) = (g \circ f)(x)$ and $L(x,1) = x$. Define $$r : M_f \to X, r([x,t]) = [L(x,t),1], r([y]) = [g(y),1] .$$ This is a well-defined continuous map (check what happens if $[y] = [x,0]$ in $M_f$). Clearly it is a retraction. For the CW case we may assume that $g$ and $L$ are cellular. Then also $r$ is cellular.
There exists a deformation of $M_f$ into $X$ (this is a homotopy $D : M_f \times I \to M_f$ such that $D(m,0) = m$ and $D(m,1) \in X$) if $f$ has a right homotopy inverse $g$.
$D$ is constructed in three steps:
a) Take the standard strong deformation retraction $\Phi :M_f \times I \to M_f$ beginning with the identity and ending with $i \circ q$, where $i : Y \to M_f$ is the canonical embedding and $q$ the canonical retraction.
b) Let $R : Y \times I \to Y$ be a homotopy such that $R(y,0) = y$ and $R(y,1) = (f \circ g)(y)$. The homotopy $i \circ R \circ (q \times id_I)$ begins with $i \circ q$ and ends with $i \circ f \circ g \circ q$.
c) Define $\Psi : Y \times I \to M_f, \Psi([y],t) = [g(y),t]$. Note that $\Psi([y],0) = [g(y),0] = [(f \circ g)(y)]$. This homotopy shifts $Y$ into $X$. The homotopy $\Psi \circ (q \times id_I)$ begins with $i \circ f \circ g \circ q$ and ends with $j \circ g \circ q$.
d) Now $D$ is obtained by composing the above three homotopies. In the CW case we get again a cellular homotopy.
We now prove that $X$ is a deformation retract of $M_f$. Take the retraction $r$ and the deformation $D$ from above. Define $$\Delta : M_f \times I \to M_f, \Delta(m) = \begin{cases} D(m,2t) & t \le 1/2 \\ (j \circ r \circ D)(m,2-2t) & t \ge 1/2 \end{cases}$$ This homotopy deforms the identity on $M_f$ to $j \circ r$. It is cellular.