1
$\begingroup$

Let $X$ and $Y$ be CW complexes, and let $f$ and $g$ be homotopic cellular maps from $X$ to $Y$; that is, $f(X^n) \subset Y^n$ and $g(X^n) \subset Y^n$, where $X^n$ denotes the $n$-skeleton of $X$. How do I show that $f$ and $g$ are cellularly homotopic (homotopic via a homotopy that is itself a cellular map)?

My attempt. Consider the relative CW complex $(X \times I, X \times \partial I)$, and let $h: f \simeq g$. We may apply the cellular approximation theorem to $h:(X \times I, X \times \partial I) \to (Y,Y)$ to get a homotopy $H: h \simeq h' \text{ rel } X \times \partial I$ where $h': X\times I \to Y$ is cellular with $h'_0=f$ and $h'_1=g$.

But something is wrong—I have not used the fact that $f$ and $g$ are cellular!

EDIT (after contemplating freakish's answer):

I got confused originally partially because of the subtle statement of the cellular approximation theorem in May's A Concise Course in Algebraic Topology:

Theorem (Cellular Approximation). Any map $f: (X,A) \to (Y,B)$ between relative CW complexes is homotopic relative to $A$ to a cellular map.

My original understanding of this result was flawed; if I applied the result as stated above to my above attempt, what I would really obtain is a homotopy $H: h \simeq h' \text{ rel } X \times \partial I$ as above but a cellular map $h':(X \times I, X \times \partial I) \to (Y,Y)$ instead (with $h'_0=f$ and $h'_1=g$ as above). This is not the same as a cellular map $h': X \times I \to Y$, as is made obvious by the following wrong proof:

False result. All maps between CW complexes are cellular.

oof. Let $f:X \to Y$ be any map. Then we may view it as a map $f:(X,X) \to (Y,Y)$, so $f$ is homotopic relative to $X$ to a cellular map by cellular approximation. That is, the homotopy is constant on $X$, so that $f$ is cellular.

Though, in a similar light, any map $\varphi:(X,A) \to (Y,Y)$ is (trivially) cellular as $(Y,Y)^n=(Y,Y)^0=Y$ and $\varphi((X,A)^n) \subset Y$ trivially. Which means my original attempt was pretty flawed.

Here then is a proof that works, with much detail so that I (hopefully) will understand it in the future:

Proof that works. Let $f$, $g: X \to Y$ be homotopic cellular maps, and let $h: f \simeq g$. We wish to find a cellular homotopy $h': f \simeq g$; that is, a homotopy $h': X \times I \to Y$ between cellular maps that is a cellular map itself. That is, we require that $h'$ sends the $n$-skeleton $X^n \times \partial I \cup X^{n-1} \times I$ of $X \times I$ into $Y^n$. Since cellular homotopies are between cellular maps, $h'(X^n \times \partial I) \subset Y^n$ automatically, so it suffices to show that $h'(X^{n-1} \times I) \subset Y^n$.

Regard $h$ as a map of relative CW complexes $(X \times I, X^n \times \partial I) \to (Y, Y^n)$. Then the cellular approximation theorem gives us a homotopy $H: h \simeq h^n \text{ rel } X^n \times \partial I$ such that $h^n: (X \times I, X^n \times \partial I) \to (Y, Y^n)$ is cellular with $h^n_0|X^n=f|X^n$ and $h^n_1|X^n=g|X^n$.

Since $h^n$ is cellular, it takes the relative $n$-skeleton $X^n \times \partial I \cup X^{n-1} \times I$ of $(X \times I, X^n \times \partial I)$ into $Y^n$. Thus $h^n$ defines the desired homotopy $h'$ on $X^n$ for each $n \geq 1$, and we may take the colimit to obtain $h'$. $\square$

Though it is nicer to just use Hatcher's version of the cellular approximation theorem.

$\endgroup$
0
2
$\begingroup$

But something is wrong—I have not used the fact that $f$ and $g$ are cellular!

You did, it's just a hidden requirement of the cellular approximation theorem. Recall:

Cellular Approximation Theorem: Every map $f:X\to Y$ of CW complexes is homotopic to a cellular map. If $f$ is already cellular on a subcomplex $A\subseteq X$ the homotopy may be taken to be stationary on $A$.

(see Allen Hatcher "Algebraic Topology", Theorem 4.8)

So in order to get that $h'_0=f$ and $h'_1=g$ you need to know that $H$ can be chosen to be stationary on $X\times \partial I$, which is a subcomplex. And this can be done if $h$ restricted to that subcomplex is cellular. And that requires $f$ and $g$ to be cellular.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.