Why is $H_k(X^n) = H_k(X)$, $k < n$, where $X^n$ is the $n$-skeleton of the CW-complex $X$?

Question

Why is $H_k(X^n) = H_k(X)$, $k < n$, where $X^n$ is the $n$-skeleton of the CW-complex $X$?

I am probably overlooking something trivial. I tried using the fact that $H_k(X^n,\emptyset)\cong H_k(X^{n-1},\emptyset) \cong \cdots \cong H_k(X^{k+1},\emptyset)$ but this appears to be a deadend.

Answer 1

You went in the wrong direction. Use that $H_k(X^n,\emptyset) \cong H_k(X^{n+1},\emptyset) \cong H_k(X^{n+2},\emptyset) \cong \cdots$, together with the usual limiting argument based on the fact that every chain is supported in some finite dimensional skeleton.

Answer 2

Let $Y$ be $X$ with a open ball from one of its $m$ cells removed where $m>n$, we show that $H_k(X)\cong H_k(Y)$ for $k<m$. By induction this gives your desired result. The isomprphism follows from the long exact sequence of a pair once we show that $H_k(X,Y)=0$. This follows from excision, since it is isomorphic to $H_k(B^m,S^{m-1})$ where $B^m$ is a $m$ dimensional ball and $S^{m-1}$ its boundary. Finally $H_k(B^m,S^{m-1})=0$ for $k<m$.