When we start reading cellular homology,we begin with this basic but important theorem:
Theorem: Let $X$ be a CW-complex and $X^m$ denote the $m$-th skeleton of its CW structure.Then we have the following:
$1. H_n(X^n,X^{n-1})\cong\bigoplus\limits_{\alpha \text{ : # of $n$ cells in $X$ }} \mathbb Z$ and $H_k(X^n,X^{n-1})=0$ for all $k\neq n$.
$2. H_k(X^n)=0$ for $k>n$ .
$3.$ The inclusion map $i:X^n\hookrightarrow X$ induces an isomorphism on the homology level $i_\ast:H_k(X^n)\xrightarrow{\cong} H_k(X)$ provided $k<n$.So we can say $H_k(X^{k+1})\cong H_k(X^{k+2})\cong\dots\cong H_k(X^{k+m})\cong H_k(X)$
The proof of this theroem is not what I am worried about but what I am worried about is how to remember this theorem:
One way I could find for remembering is as follows:
$1.$ As far as the relative homologies are concerned $X^n/X^{n-1}$ has no higher dimensional holes,as there is no higher dimensional cell in $X^n$.On the other hand,$X^n/X^{n-1}$ collapses all the lower dimensional cells ,which are $\leq n-1$ dimensional and hence $H_k(X^n,X^{n-1})=0$ for all $k\neq n$ and if we investigate the case $k=n$ we will find out that $X^n/X^{n-1}$ is homotopically equivalent to $\bigvee_{\alpha} S^n$ ,a wedge sum of $n$-spheres where the index $\alpha$ runs over the number/cardinality of $n$-cells in $X$ and hence by Mayer-Vietoris $H_n(X^n,X^{n-1})\cong \bigoplus_\alpha \mathbb Z$ with the direct sum running over the same indexing set indexed by $\alpha$ as the wedge sum.This allows us to conclude that $H_k(X^n,X^{n-1})=\begin{cases} \bigoplus_\alpha \mathbb Z \text{ if } k=n \\0 \text{ if } k\neq n\end{cases}$.
$2.$ For this part I want to think in this way,that $X^n$ has no $\geq n+1$ dimensional/ higher dimensional holes because there is no higher dimensional cells.So,it becomes evident from the intuitive interpretation of Homology.
$3.$ I want to remember the third part in the following way.To find the $k$th homology we need to know the cell structure only up to $k+1$ th dimension because in the chain level we will need just the maps $\partial_k$ and $\partial_{k+1}$ in order to get hold of the homology group $H_k(X)=\ker(\partial_{k})/\operatorname{Im}(\partial_{k+1})$,the higher skeletons have no roles or contribution in this computation,hence this part can also be fitted to our intuition.
This is how I like to remember this theorem.And I think a good example to keep in mind for a quick verification is the example of $X=\mathbb T^2$ the torus,the homologies of its skeletons and the relative homologies of one skeleton by the preceeding skeleton.