I'm reading through Hatcher and Lemma 2.34 states the inclusion $X^n \to X$ induces an isomorphism from $H_k(X^n)$ to $H_k(X)$ for $k < n$. I'm reading through the proof of the infinite-dimensional case, and I'm currently stuck on the following reduction of the problem.
If we put $(X^n,X)$ into a long exact sequence, portions of the sequence look like $H_{k+1}(X,X^n) \to H_k(X^n) \to H_k(X) \to H_k(X,X^n)$. If we can show that $H_k(X,X^n) = 0$ for $k \leq n$, we would prove the result.
Hatcher then claims it is sufficient to show that if $X^n$ is a point, then $H_k(X) = 0$. I understand that this would imply $H_k(X,X^n)$ would be trivial, but I don't see why we can make such a big assumption on the $n-$skeleton of $X$. He also claims that $H_k(X,X^n) \cong \tilde{H}_k(X/X^n)$ but I don't know why this is relevant.
The claim that it is sufficient to show $\tilde H_{\le n}(Y)=0$ if $Y^n$ is a point is directly linked to the isomorphism $H_k(X,X^n)\cong \tilde H_k(X/X^n)$. We want to show that $H_k(X,X^n)=0$ for $k\le n$. By the isomorphism, it suffices to show that $\tilde H_k(X/X^n)=0$ for $k\le n$. Now $X/X^n$ is homeomorphic to a CW complex whose $n\text{-skeleton}$ is a point (see this post).
The isomorphism $H_k(X,X^n)\cong \tilde H_k(X/X^n)$ comes from $(X,X^n)$ being a good pair (see page 124).