I'm trying to solve a problem of Hatcher's Algebraic Topology:

"Use cellular approximation to show that the n-skeletons of homotopy equivalent CW-complexes without cells of dimension n+1 are also homotopy equivalent."

Using cellular approximation I can get a cellular map which is also a homotopy equivalence between the two CW-complexes. Should I prove that the restriction of it to the n-skeleton is still a homotopy equivalence? Is there any theorem about how to get a homotopy equivalence after a restriction?

This is what I did so far:

Let $f:X\to Y$ be the homotopy equivalence and $\widetilde{f}$ its homotopic cellular map with $H:X\times [0,1]\to Y$ the homotopy between them. I know that $X\times [0,1]$ inherits a CW-structure from $X$ and therefore, I can use again cellular approximation to get a cellular homotopy $\widetilde{H}$, homotopic to $H$. By assuming $X_m=X_{m+1}$ and $Y_m=Y_{m+1}$, I get $$(X\times [0,1])_{m+1}=\bigcup_{p+q=m+1}X_p\times [0,1]_q= X_{m+1}\times \{0\}\cup X_{m+1}\times \{1\} \cup X_{m}\times [0,1]= X_{m}\times [0,1]$$ then the restriction of $\widetilde{H}$ to $(X\times [0,1])_{m+1}$ is still a homotopy between two maps $X_m\to Y_m$. Hence I have two maps that in some way are homotopic to $f$. Am I going somewhere?


1 Answer 1


Your idea is absolutely correct.

You have maps $f : X \to Y$ and $g : Y \to X$ which are homotopy inverse to each other. Choose cellular maps $f' : X \to Y$ and $g' : Y \to X$ homotopic to $f$ and $g$, respectively. Then they are also homotopy inverse to each other. We claim that the restrictions $f_n' : X^n \to Y^n$ and $g_n' : Y^n \to X^n$ of $f'$ and $g'$ are homotopy inverse to each other.

There exists a homotopy $H : X \times I \to X$ from $g' f'$ to $id$. Its restriction to $X \times \{0,1\}$ is cellular, thus it is homotopic rel. $X \times \{0,1\}$ to a cellular map $H' : X \times I \to X$. We have $H'_0 = g' f'$ and $H'_1 = id$. Since $X^n \times I \subset (X \times I)^{n+1}$ and $X^{n+1} = X^n$, we get $$H'(X^n \times I) \subset X^{n+1} = X_n .$$ This shows that $g'_n f'_n \simeq id$. That $f'_n g'_n \simeq id$ is similar.


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