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?

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    $\begingroup$ You are right. If you restrict the antipodal action on $S^n$ to the equatorial sphere $S^{n-1}$, it gives rise to the antipodal action on this sphere. Therefore in the quotient, the equatorial sphere becomes and $\mathbb{RP}^{n-1}$, to which you are attaching a single $n$-ball. In your set-up, $R$ deformation retracts to the quotient of the equatorial sphere. $\endgroup$ Mar 14, 2011 at 18:23
  • $\begingroup$ @Jim: Yes, that was the basis for my intuition. However, I was not sure it was a valid argument: consider the universal cover $\mathbb{R} \to S^1$. $\mathbb{R}$ retracts to a point, $S^1$ doesn't. What's the essential difference between that and $S^{n-1} \times B^1 \to R$ that makes it work? $\endgroup$
    – Zhen Lin
    Mar 14, 2011 at 18:29
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    $\begingroup$ You have to make sure that you can do the deformation retraction equivariantly. That is, if two points are antipodal, they get mapped to antipodal points at all stages of the deformation retraction. $\endgroup$ Mar 14, 2011 at 19:27
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    $\begingroup$ I think it would be much easier to simply compute the cellular homology of $\mathbb{R}\mathbb{P}^n$ (which would involve computing the boundary homomorphism $\partial_i:(\mathbb{R}\mathbb{P}^{i+1},\mathbb{R}\mathbb{P}^{i})\to (\mathbb{R}\mathbb{P}^{i},\mathbb{R}\mathbb{P}^{i-1})$ for all $0\leq i\leq n-1$ (where $\mathbb{R}\mathbb{P}^{-1}=\emptyset$ by convention) unlike in the case of complex projective space where the groups constituting the relevant cellular chain complex completely determine the boundary homomorphism in all dimensions) ... $\endgroup$ Dec 22, 2011 at 8:26
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    $\begingroup$ ... You can then argue that the cellular homology groups and the simplicial homology groups are isomorphic in the case of a triangulable CW-complex. $\endgroup$ Dec 22, 2011 at 8:26

1 Answer 1


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.

  • $\begingroup$ Also, I believe the author has provided enough information to compute the groups using his method. The maps he mentioned induce vertical chain maps on the homology groups for each term in the Meyer Vietoris sequence. To apply the sequence, you should assume $M$ and $L$ overlap a bit. Then $L\cap M\simeq S^{n-1}\coprod S^{n-1}$ and $M\simeq S^{n-1}$. Examine what these maps do on the level of homology. Use the fact that the quotient map $S^{n-1}\simeq M \to R\simeq \mathbf{R}P^{n-1}$ $\endgroup$
    – user113529
    Dec 11, 2013 at 1:20

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