Try to generalize a problem in Hatcher: finite vs. infinite CW-complexes While solving a problem in Hatcher I got this doubt in my mind,
In the 2nd chapter (Homology) Hatcher asked us to prove the following question...

If $X$ is a finite dimensional CW-complex then,
a) If $X$ has dimension n, then  $H_i(X)=0$ for $i>n$ and $H_n(X)$ is a free group.
b) $H_n(X)$ is free with basis in bijective correspondence with the $n$-cells if there is no $n+1$ and $n-1$ cell.
c) If $X$ has $k$ $n$-cells, then $H_n(X)$ is generated by at most $k$ elements.

While proving the first part I've used induction over $n$ and for the rest of the proof I used the fact that $H_n(X^{n+1})=H_n(X^k)$ for $k>n$ and since $X$ is a finite dimensional CW-complex so $H_n(X^{n+1})=H_n(X)$, so finiteness property of the CW-structure plays a key role through out my solution for this problem.
Now my question is this:

Are these results ,b) and c) still hold if I consider $X$ a CW-complex only?

For this my solution won't work anymore. Can anyone provide me some counter example if it is not true, or some direct proof of this fact?
 A: You are right that $H_n(X^{n+1})\approx H_n(X^m)$ for all $m>n\ge 0$, and this isomorphism is induced by inclusion. If $X$ is finite-dimensional, then it follows easily that $H_n(X^{n+1})=H_n(X)$. However, finite-dimensionality is not the reason for this equality, as it also holds in case that $X$ is infinite-dimensional. This is because every chain $\gamma$ is a sum of finitely many simplices and is thus contained in some skeleton $X^m$. That means $[\gamma]\in H_n(X)$ is in the image of $H_n(X^m)$, but this group is the image of $H_n(X^{n+1})$. By the same argument, if an $n$-chain $\gamma$  bounds a chain $\beta$, then $[\gamma]$ is zero already in some $H_n(X^m)$, thus is must be zero in $H_n(X^{n+1})$.
Note that we also have isomorphisms $i_*:H_k(X^n)\to H_k(X^{n+1})$ whenever $k>n+1>0$. We know that $H_k(X^0)=0$ for all $k>0$, and if we assume by induction that $H_k(X^n)=0$ for all $k>n\ge 0$, it follows that $H_k(X^{n+1})=0$ whenever $k>n+1$.
Combining the long exact sequences for the pairs $(X^{n+1},X^n)$ and defining $d_n=\partial_n j_n$, we get a diagram
 (image from Algebraic Topology by Allen Hatcher, page 139)
where $(d_n)_n$ forms a chain complex, called the cellular chain complex. It is not difficult to show that 
$H_n(X)\approx \text{Ker}(d_n)/\text{Im}(d_{n+1})$, so the homology groups of $X$ can be computed via this cellular chain complex.
Of course, it would be nice to have a more explicit formula for the maps $d_n$, but we can still deduce some immediate consequences from what we have so far:


*

*If $X$ has no cells in dimension $n+1$ and $n-1$, then $d_n$ and $d_{n+1}$ are zero, hence $X$ is freely generated by the $n$-cells.

*If $X$ has $k$ $n$-cells, $H_n(X)$ is generated by at most $k$ elements.


For the details, check section $2.2$ in Hatcher's book.
