A claim in cellular homology 


These pages are from Allen Hatcher's book.
In the proof of (a), I did not get how they are using 2.22. And 2.25. In the proof.
And in (b), how they conclude that if $X$ is finite dim then $H_k(X)=0$ for all $k>dim X$.
In b: I looked at the long exact sequence of $(X^n,X^{n-1})$:
$...\to H_{k+1}(X^n,X^{n-1}) \to H_k(X^{n-1})\to H_k(X^n)\to H_k(X^n,X^{n-1})\to....$.
And get that $H_k(X^{n-1}) \simeq H_k(X^n)$ for all $k>n$. So, for $k>n$:
$H_k(X^n)\simeq H_k(X^{n-1})\simeq....\simeq H_k(X^0)=0$ (since k>0).
I know also that $X$ is finie dim so there is $N$ big enough such that $=X^N$.
And I would be grateful if you can explain the concept of (c) when $X$ is infinite dim (the direct approach).
Highly appreciate every clarification.
 A: Let's go through your concerns.

*

*In part (a), we want to show something about $H_{k}(X^{n}, X^{n-1}).$ The pair $(X^{n}, X^{n-1})$ is a CW pair, hence is a good pair, so we can apply Proposition 2.22 and conclude that
$$H_{k}(X^{n}, X^{n-1}) \cong \tilde{H}_{k}(X^{n}/X^{n-1}).$$ What is $X^{n}/X^{n-1}?$ Well, $X^{n}$ is a bunch of $n$-cells attached by their boundaries to $X^{n-1}$, so when we quotient out by $X^{n-1}$ all of the boundaries of these $n$-cells get crushed to a point. So, $X^{n}/X^{n-1}$ is a wedge of $n$-spheres, one for each $n$-cell of $X$.
We then have
$$\tilde{H}_{k}(X^{n}/X^{n-1}) = \tilde{H}_{k}(\bigvee_{n\text{-cells of } X}S^{n}).$$ By Corollary 2.25 we have
$$\tilde{H}_{k}(\bigvee_{n\text{-cells of } X}S^{n}) \cong \bigoplus_{n\text{-cells of } X} \tilde{H}_{k}(S^{n}).$$ From here the result should be clear.

*From what you wrote in the question it seems like you've got part (b), so I won't write about it.

*For the infinite dimensional case in part (c), the direct approach goes as follows:  take a singular $k$-chain in $X$. This is a finite linear combination of maps $\Delta^{k} \to X$, each of which has compact image because $\Delta^{k}$ is compact. So, the image of the singular $k$-chain is compact as well, hence by Proposition A.1 only meets finitely many cells of $X$. Then, there exists some finite $m$ such that $X^{m}$ contains all these finitely many cells, and so the singular $k$-chain is also a singular $k$-chain in $X^{m}$. So, if we take a $k$-cycle in $X$, there is some finite $m$ such that it is also a $k$-cycle in $X^{m}$.

Fix $n \geq k$. Without loss of generality we can take $m \geq n$.
Let $Y = X^{m}$. Then, $Y$ is a finite dimensional CW complex, so by
the finite dimensional case our $k$-cycle is homologous to a
$k$-cycle in $Y^{n}$. But the $n$-skeleton of $Y^{n}$ is just the
$n$-skeleton of $X$ (because $Y = X^{m}$ with $m \geq n$), so it
follows that our $k$-cycle is homologous to a $k$-cycle in $X^{n}$.
This proves surjectivity. The argument for injectivity is similar,
so I leave it to you.
