Consider the $n$-sphere $S^n$ and the real projective space $\mathbb{RP}^n$. There is a universal covering map $p: S^n \to \mathbb{RP}^n$, and it's clear that it's the coequaliser of $\mathrm{id}: S^n \to S^n$ and the antipodal map $a: S^n \to S^n$. At first I thought I might use this fact and invoke abstract nonsense to compute the homology groups of $\mathbb{RP}^n$ but then I realised that it didn't give the correct answer. The rubric of the question suggests the following approach:

- Cut $S^n$ along two parallels on opposite sides of the equator to obtain two subspaces $L$, homeomorphic to two disjoint copies of the closed ball $B^n$, and $M$, homeomorphic to the closed cylinder $S^{n-1} \times B^1$.
- The antipodal map sends $L$ to $L$ and $M$ to $M$, and so they are both double covers of their images $Q$ and $R$ (resp.) in $\mathbb{RP}^n$. Clearly $Q$ is homemorphic to a single copy of $B^n$, and with some thought it's seen that $R$ is a $n$-dimensional generalisation of the Möbius strip.
- Use the fact that the $a$ and $p$ act nicely on $L$ and $M$ to obtain chain maps between the Mayer—Vietoris sequences for $S^n = L \cup M$ and $\mathbb{RP}^n = Q \cup R$.

**Edit**: Following Jim Conant's comment about deforming $R$ into $\mathbb{RP}^{n-1}$, I managed to compute the homology of $\mathbb{RP}^n$ by induction (on $n$, of course). However, I'm not convinced this was the intended solution, as the question explicitly refers to the fact that any nice simplicial map $f: K \to P$ sending subcomplexes $L, M$ into $Q, R$ (resp.), where $K = L \cup M$ and $P = Q \cup R$, induces a chain map $f_*$ between the Mayer—Vietoris sequences
$$\cdots \to H_{r+1} (K) \to H_r (L \cap M) \to H_r(L) \oplus H_r(M) \to H_r(K) \to H_{r-1} (L \cap M) \to \cdots$$
and
$$\cdots \to H_{r+1} (P) \to H_r (Q \cap R) \to H_r(Q) \oplus H_r(R) \to H_r(P) \to H_{r-1} (Q \cap R) \to \cdots$$
but I have not used this fact anywhere in my solution. Have I made a mistake somewhere?