It is well known that if $M$ is a compact orientable $n$-dimensional manifold, then $[M, \mathbb{S}^n] \cong \mathbb{Z}$, i.e the maps are classified by their degree.

What is known about $[M, \mathbb{RP^n}]$ under the same hypotheses?

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    $\begingroup$ To have $[M, S^n]\cong \mathbb Z$ you need to assume that M is orientable. $\endgroup$ Nov 17, 2019 at 21:30
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    $\begingroup$ Since the infinite real projective space classifies first cohomology with $\mathbb{Z}/2$ coefficients and the $n$ dimensional projective space is its $n$ skeleton you get that you have a canonical surjection to the first cohomology of your manifold. The kernel can be described using obstruction theory. $\endgroup$
    – ThorbenK
    Nov 17, 2019 at 21:32

1 Answer 1


This seems to have been worked out in the 1960s by Paul Olum, see Section 1 of

Olum, P., Cocycle formulas for homotopy classification; maps into projective and lens spaces, Trans. Am. Math. Soc. 103, 30-44 (1962). ZBL0135.23203.

Briefly summarizing, two based maps $f,g:M\to \mathbb{R}P^n$ are based homotopic if they agree on the generators of $H^1(\mathbb{R}P^n;\mathbb{Z}/2)$ and $H^n(\mathbb{R}P^n;\mathbb{Z}^w)$ (twisted cohomology; this has to be interpreted in terms of the difference homomorphism $(f-g)^*$ in twisted coefficients, described in the Appendix), and a certain difference cocycle in $C^n(\mathbb{R} P^n,x_0;\mathbb{Z}/2)$ represents zero in $H^n(\mathbb{R}P^n,x_0;\mathbb{Z}/2)$. The result for free homotopy classes is then deduced from this.

So, secondary operations are needed to describe the answer, but this seems to be one of the few cases where this can be done explicitly.


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