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A friend gave me the tip of using cellular approximation. I think that the idea is to classify only cellular maps since any map is homotopic to a cellular one. So, I wrote down what a cellular map $\mathbb{R}P^2\to\mathbb{C}P^2$ should look like: $$sk_{0}(\mathbb{R}P^2)=\ast\to\ast=sk_{0}(\mathbb{C}P^2)$$ $$sk_{1}(\mathbb{R}P^2)=S^{1}\to\ast=sk_{1}(\mathbb{C}P^2)$$ $$sk_{2}(\mathbb{R}P^2)=\mathbb{R}P^2\to S^{2}=sk_{2}(\mathbb{C}P^2)$$ Since there is only a unique map from any space to a point, it comes down to classifying maps $\mathbb{R}P^2\to S^{2}$ , but I am lost. If it was the other way around, that is computing $[S^{2},\mathbb{R}P^2]$, then sure, we could use the real Hopf fiber bundle and compute $\pi_2(\mathbb{R}P^2)$, but with the maps going this way I have no idea. Any tips?

Henry Badhead
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  • What tools do you have available? Do you know about Eilenberg-Mac Lane spaces and representability of ordinary cohomology, for instance? – Ben Steffan Sep 29 '24 at 12:37
  • @Ben Steffan I know vaguely about it, but we did not prove it. However, I would still be interested in hearing how one would use it. As far as tools go, roughly I have: covering theory, cellular homology, homotopy excision theorem, Freudenthal, basics about co/fibrations, UCTs and Poincaré duality – Henry Badhead Sep 29 '24 at 12:42
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    Since you're asking for a tip: Find a suitable cofibre sequence involving $\mathbb{R}\mathrm{P}^2$ and use the (Baratt-)Puppe exact sequence. – Ben Steffan Sep 29 '24 at 12:56
  • @Ben Steffan Thank you very much, I've looked the Puppe sequence up and thought a bit about it, but do not seem to get it. So, if you have the time to write down the solution, or some additional hints, like which cofibre sequence, I would be very thankful to you! – Henry Badhead Sep 29 '24 at 17:54
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    By the way, the Hopf theorem is equidimensional (and degree assumes orientability). – Ted Shifrin Sep 29 '24 at 21:04
  • @Ted Shifrin I had similar doubts, hence I edited several times the question. However, regarding equidimensionality, the spaces that I am considering in the end are indeed equidimensional, $\mathbb{R}P^2$ and $S^2$, via the argument with cellular approximation. Regarding orientability: isn't it the case that we still have $\mathbb{Z}/2$-degrees for non-orientable manifolds? As they are defined here https://planetmath.org/DegreeMod2OfAMapping – Henry Badhead Oct 01 '24 at 07:56
  • Yes, you have mod $2$ degree, but not the Hopf theorem. – Ted Shifrin Oct 01 '24 at 13:34
  • @Ted Shifrin Ahhh Ooops! Thank you very much! This is useful to keep in mind – Henry Badhead Oct 01 '24 at 14:12
  • Also shoutout to @lisan101 who also helped a lot along the way! – Henry Badhead Oct 01 '24 at 14:15

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Since you asked for it in the comments, here's two ways to compute this set, one using a Barratt-Puppe sequence and one using representability of ordinary cohomology.

Note that we have a cofibre sequence $\newcommand{\RP}{\mathbb{R}\mathrm{P}}S^1 \xrightarrow{\cdot 2} S^1 \to \RP^2$, so applying $[{{-}}, S^2]$ (I am using your reduction to $S^2$ in the codomain here, but the argument would work just as well with $\newcommand{\CP}{\mathbb{C}\mathrm{P}}\CP^2$) we obtain an exact sequence $$ \cdots \to \underbrace{[\Sigma S^1, S^2]}_{= \pi_2 S^2 \cong \mathbb{Z}} \xrightarrow{\cdot 2} \underbrace{[\Sigma S^1, S^2]}_{{= \pi_2 S^2 \cong \mathbb{Z}}} \to [\RP^2, S^2] \to \underbrace{[S^1, S^2]}_{= 0} \to \underbrace{[S^1, S^2]}_{= 0} $$ and conclude that $|[\RP^2, S^2]| = 2$; in fact, we can identify $[\RP^2, S^2]$ with the cokernel of the map $\pi_2 S^2 \xrightarrow{\cdot 2} \pi_2 S^2$ in this case so that it carries a group structure and is isomorphic to $\mathbb{Z} / 2$.

(N.b. that care must be taken in sequences like this; they are a priori exact sequences of pointed sets, not of groups. Here we got lucky.)

For proof 2, note that $[\RP^2, \CP^2] \cong [\RP^2, \CP^\infty]$ by an analogous cellular approximation argument, so since $\CP^\infty$ represents $H^2({{-}}; \mathbb{Z})$ we conclude that $[\RP^2, \CP^2] \cong H^2(\RP^2; \mathbb{Z}) \cong \mathbb{Z} / 2$, agreeing with our computation above.

(N.b. that this argument can (easily) be generalized to the Hopf-Whitney theorem, stated e.g. here).

As a last remark, one can give an explicit representative of the non-trivial class in $[\RP^2, \CP^2]$: The inclusion $\mathbb{R} \hookrightarrow \mathbb{C}$ induces a map $\RP^n \to \CP^n$ and thus also a map $\iota\colon \RP^\infty \to \CP^\infty$. One can show that $\iota^*\colon H^*(\CP^\infty) \to H^*(\RP^\infty)$ identifies with the quotient map $\mathbb{Z}[x] \to \mathbb{Z}[x] / (2x)$, so its restriction to the 2-skeleton induces a non-trivial map $H^*(S^2) \to H^*(\RP^2)$ which means it must be homotopically non-trivial.

Ben Steffan
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