Homology of CW Complex Mod Skeleton I'm trying to figure out a general approach to computing the homology groups for a space $X/X^m$ where $X$ is a CW complex of dimension $n$, and $X^m$ is its $m$-skeleton where $m<n$.  I'm aware that the homology groups for $X$ are derived from the fact that $(X^{k+1}, X^k)$ are good pairs for all $k$.  
I'm in particular trying to do this with quotient space $\mathbb{RP}^n/\mathbb{RP}^m$ where we use the standard cell decomposition of $\mathbb{RP}^n$.  I'm aware of the exact sequence
$$
\ldots \to \widetilde{H}_k\left (\mathbb{RP}^m\right ) \to \widetilde{H}_k\left (\mathbb{RP}^n \right ) \to \widetilde{H}_k\left (\mathbb{RP}^n/\mathbb{RP}^m\right ) \to \widetilde{H}_{k-1}\left (\mathbb{RP}^m\right ) \to \ldots
$$
where the induced maps are (from left to right) given by inclusion, quotient, and then boundary.  I'll mention here that I'm more or less comfortable with the process of diagram chasing, but I struggle with inferring what the boundary and attachment maps are when it comes to various constructions of spaces.  
BTW, this is Hatcher exercise 2.2.19.
 A: There are a few things you can say in general. First, note that $H_i(X^m) = 0$ if $i> m$ and $H_i(X^m) \to H_i(X)$ is an isomorphism for $i < m$. From these and the long exact sequence of the pair it follows that $H_i(X, X^m)\cong H_i(X)$ when $i > m+1$, and is $0$ if $i < m$. For the remaining two values we have an exact sequence
$$ 0 \to H_{m+1}(X) \to H_{m+1}(X, X^m) \to H_m(X^m) \stackrel{\varphi}{\to} H_m(X) \to H_m(X, X^m) \to 0 $$
where $\varphi$ is the homomorphism induced by inclusion. Then it follows that $H_m(X, X^m) \cong coker(\varphi)$, and $H_{m+1}(X, X^m)$ is an extension of $ker(\varphi)$ by $H_{m+1}(X)$. I'm not sure if it's possible to say anything else without considering a specific case.

So let's consider the specific case of $\mathbb{RP}^m \subset \mathbb{RP}^n$ for $m < n$. Using the above, try to work out the computation for yourself before reading the rest of my answer.
Recall that $H_i(\mathbb{RP}^n) \cong \mathbb{Z}/2$ if $1\leq i < n$ is odd, $\mathbb{Z}$ if $i=0$ or if $i=n$ and $n$ is odd, and $0$ otherwise. This means we have to consider both of the cases $m$ odd and $m$ even.
If $m$ is even then $H_m(\mathbb{RP}^m)=0$, so for $i=m, m+1$ we simply have $H_i(\mathbb{RP}^n, \mathbb{RP}^m) \cong H_i(\mathbb{RP}^n)$.
If $m$ is odd then $H_m(\mathbb{RP}^m) \cong \mathbb{Z}$ and $H_{m+1}(\mathbb{RP}^n) = 0$, and our sequence takes the form
$$0 \to H_{m+1}(\mathbb{RP}^n, \mathbb{RP}^m) \to \mathbb{Z} \stackrel{\varphi}{\to} \mathbb{Z}/2 \to H_m(\mathbb{RP}^n, \mathbb{RP}^m) \to 0 $$
Now we need a geometric argument to understand $\varphi$. When we consider the inclusion $\mathbb{RP}^m \subset \mathbb{RP}^n$, the top $m$-cell of the subspace gets an $(m+1)$-cell attached to it in the larger space via a map of degree $2$, and from the cellular chain complex you see that this $m$-cell that generates $H_m(\mathbb{RP}^m)\cong \mathbb{Z}$ also generates $H_m(\mathbb{RP}^n) \cong \mathbb{Z}/2$. In other words $\varphi$ is surjective, and therefore $H_m(\mathbb{RP}^n, \mathbb{RP}^m) = 0$ and $H_{m+1}(\mathbb{RP}^n, \mathbb{RP}^m)\cong ker(\varphi)\cong \mathbb{Z}$.
(Note that this result is consistent with the fact that $\mathbb{RP}^{m+1}/\mathbb{RP}^m \cong S^{m+1}$.)
