Simplicial homology of real projective space by Mayer-Vietoris

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:

  1. 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 $S^{n-1} \times B^1$.
  2. 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.
  3. 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?

Answer

This is not really an answer to your question, and I’m not sure how your original attempt went, but one has to be careful when taking homology of colimit diagrams of $spaces$. Homology commutes (up to natural iso) with colimits in the category of chain complexes. This is not necessarily true of the composite functor $Top \overset{S_{*}}{\to} Ch_{+} \overset{H}{\to} Ab_{gr}$. Here $S_{*}$ is the functor which assigns to each space, its singular chain complex and $Ab_{gr}$ is the category of graded abelian groups. This is true, however, for $filtered$ colimits. In particular, if $X=\bigcup_{p}X_{p}$, then
$$H_{*}(X)\simeq \lim_{\leftarrow}H_{*}(X_{p}).$$
What I’m saying is, more precisely, that the coequalizer diagram $S^{n}\substack{ \overset{\alpha}\to \\ \to} S^{n}$ is not a filtered colimit. You cannot therefore conclude that the coequalizer of $H_{*}(S^{n})\substack{\to \\ \to}H_{*}(S^{n})$ is isomorphic to the homology of $\mathbf{R}P^{n}$. Maybe you already knew this, but I was just pointing it out if you had not considered the point.

Attribution
Source : Link , Question Author : Zhen Lin , Answer Author : Community

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