The homotopy set $[\mathbb{R}P^{n+1},S^n],\,n\ge3$

Question

Question (courtesy of @BenSteffan): What is the homotopy set $[\mathbb{R}P^{n+1},S^n],\,n\ge 3$?

[This entire post takes place in the pointed category.] Note that there is, for any $n\ge1$, a cofiber sequence $$S^n\rightarrow\mathbb{R}P^n\rightarrow\mathbb{R}P^{n+1}\rightarrow S^{n+1}\rightarrow\Sigma\mathbb{R}P^n\rightarrow\dotsc.$$ This induces an exact sequence of pointed sets $$\dotsc\rightarrow[\Sigma\mathbb{R}P^n,S^2]\rightarrow[S^{n+1},S^n]\rightarrow[\mathbb{R}P^{n+1},S^n]\rightarrow[\mathbb{R}P^n,S^n]\rightarrow[S^n,S^n],$$ where furthermore the group $[S^{n+1},S^n]=\pi_{n+1}(S^n)=\mathbb{Z}/2\mathbb{Z}$ acts transitively on the fibers of the restriction map $[\mathbb{R}P^{n+1},S^n]\rightarrow[\mathbb{R}P^n,S^n]$. Now, Hopf's Theorem states that $[-,S^n]$ corepresents $H^n(-;\mathbb{Z})$ on $\le n$-dimensional CW-complexes, so the latter map is identified with the pullback $H^n(\mathbb{R}P^n)\rightarrow H^n(S^n)=\mathbb{Z}$. Now, consider two cases:

• If $n$ is odd, then $\mathbb{R}P^n$ is orientable, so the final map is $2\cdot\colon\mathbb{Z}\rightarrow\mathbb{Z}$ (since $S^n\rightarrow\mathbb{R}P^n$ is a double cover). In particular, this map is injective, so exactness yields that we have a surjection $\pi_{n+1}(S^n)\twoheadrightarrow[\mathbb{R}P^{n+1},S^n]$, so the homotopy set in question has $1$ or $2$ elements depending on whether the composite $\mathbb{R}P^{n+1}\rightarrow S^{n+1}\rightarrow S^n$ of the map collapsing $\mathbb{R}P^n$ and the $(n-2)$-fold suspension of the Hopf map $\eta\colon S^3\rightarrow S^2$ is non-trivial. It is not clear to me whether it is.

• If $n$ is even, $\mathbb{R}P^n$ is non-orientable, so the final map is the null map $\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}$. Thus, exactness yields that the restriction map $[\mathbb{R}P^{n+1},S^n]\twoheadrightarrow[\mathbb{R}P^n,S^n]=\mathbb{Z}/2\mathbb{Z}$ is surjective. Furthermore, $\pi_{n+1}(S^n)=\mathbb{Z}/2\mathbb{Z}$ acts transitively on its fibers, so the homotopy set in question has between $2$ and $4$ elements.

From this point, I cannot make any more progress. It's worth noting that the analogous question for $n=2$ can also be posed, in which case I know the answer. Namely, the projection $S^3\rightarrow\mathbb{R}P^3$ induces a bijection $[\mathbb{R}P^3,S^2]\stackrel{\sim}{\rightarrow}[S^2,S^2]=\mathbb{Z}$, which can be shown by constructing enough elements and using the above exact sequence. However, such an approach seems less feasible for $n\ge3$ since the analogous map $[\mathbb{R}P^{n+1},S^n]\rightarrow[S^{n+1},S^n]=\pi_{n+1}(S^n)=\mathbb{Z}/2\mathbb{Z}$ now fails to detect the image of $[S^{n+1},S^n]\rightarrow[\mathbb{R}P^{n+1},S^n]$ since the composite $S^{n+1}\rightarrow\mathbb{R}P^{n+1}\rightarrow S^{n+1}$ is a degree $2$ map and $\mathbb{Z}/2\mathbb{Z}$ is $2$-torsion.

Answer 1

$\newcommand{\Sq}{\operatorname{Sq}} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} $

A general theorem of Steenrod

The homotopy set $[X, S^n]$, where $X$ is a complex of dimension $n+1$, was described by Steenrod in the paper where he introduced the Steenrod operations:

Steenrod, N. Products of cocycles and extensions of mappings. Ann. of Math. (2) 48 (1947), 290–320.

A more recent account is given in Chapter 14 of the book Cohomology operations and applications in homotopy theory by Mosher and Tangora. The result is as follows:

Theorem (Steenrod): Let $X$ be a complex of dimension $n+1$, where $n \geq 3$. Then there is a bijection $$ [X, S^n] \cong H^n(X; \Z) \times \operatorname{Coker} \phi, $$ where $\phi$ is the following composition: $$ H^{n-1}(X; \Z) \xrightarrow{\mod 2} H^{n-1}(X; \Z_2) \xrightarrow{ \Sq^2 } H^{n+1}(X; \Z_2) $$

The idea of the proof given in Mosher-Tangora is to work out the first two stages of a Postnikov tower for $S^n$. The first stage is $K(\Z,n)$, and the second stage is obtained by killing a map $$ K(\Z,n) \to K(\Z_2, n+2) $$ that represents the operation $\Sq^2 \circ \rho$, where $\rho$ is reduction mod $2$.

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Computations for $X = \R P^{n+1}$

To deduce the answer for $X = \R P^{n+1}$, one needs to work out the map $\phi$. I could not find a reference for the end result. Hopefully, someone can check that I didn't make any mistakes in the computations.

The computation splits into two cases:

$n$ is even

This case is easy: we have $H^{n-1}(\R P^{n+1}; \Z) = 0$, hence $$ [\R P^{n+1}, S^n] \cong H^{n}(\R P^{n+1}; \Z) \times H^{n+1}(\R P^{n+1}; \Z_2) \cong \Z_2 \times \Z_2 $$

$n$ is odd

For $n$ odd, we have $H^{n-1}(\R P^{n+1}; \Z) \cong \Z_2$. Moreover, by looking at the exact sequence induced by the exact sequence of coefficients $$ 0 \to \Z \xrightarrow{\times 2} \Z \xrightarrow{\mod 2} \Z_2 \to 0, $$ one sees that the reduction map $H^{n-1}(\R P^{n+1}; \Z) \to H^{n-1}(\R P^{n+1}; \Z_2)$ is an isomorphism.

It remains to figure out $\Sq^2$ on $H^{n-1}$. The computation is similar to the case of $\mathbb{C}P^n$ which is worked out in this answer. Let $x \in H^1(\R P^{n+1}; \Z_2)$ be the generator. Then $\Sq(x) = x + x^2$. Since $\Sq$ is a map of rings, we have $$ \Sq(x^k)= \Sq(x)^k = (x+x^2)^k = \sum_{j=0}^k \binom{k}{j} (x^2)^j x^{k-j} =\sum_{j=0}^k \binom{k}{j} x^{k+j}. $$ Here, the binomial coefficients are reduced mod $2$. Now $\Sq^2$ corresponds to the term $j=2$ in the last sum, hence $$ \Sq^2 x^k = \binom{k}{2} x^{k+2}. $$ Now $\binom{k}{2}$ is even when $k =0,1$ mod $4$ and odd when $k = 2,3$ mod $4$. Hence we conclude $$ \Sq^2 x^k = \begin{cases} 0 &\text{if } k =0,1 \mod 4 \\ x^{k+2} &\text{if } k= 2,3 \mod 4 \end{cases} $$

The case we are interested in is $k = n-1$ where $n$ is odd, so we obtain: $$ \Sq^2 x^{n-1} = \begin{cases} 0 &\text{if } n = 1 \mod 4 \\ x^{n+1} &\text{if } n = 3 \mod 4. \end{cases} $$ Putting this all together, and assuming my computations are correct, we conclude: $$ [ \R P^{n+1}, S^n ] \cong \begin{cases} \Z_2 &\text{if } n = 1 \mod 4, \\ 0 &\text{if } n = 3 \mod 4. \end{cases} $$