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I want to calculate $[\mathbb{S}^n, \mathbb{RP}^{n-1}]$. I know that there is a correspondence $[\mathbb{S}^n, \mathbb{RP}^{n-1}] \cong \pi_n(\mathbb{RP}^{n-1})/\pi_1(\mathbb{RP}^{n-1})$ where I mean the quotient by the $\pi_1(\mathbb{RP}^{n-1})$ action. I can calculate $\pi_n(\mathbb{RP}^{n-1})$, but I don't know which is the action.

My question then is: which is the $\pi_1(\mathbb{RP}^{n-1})$-action on $\pi_n(\mathbb{RP}^{n-1})$?

CNS709
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1 Answers1

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I think the easiest description is the following:

The inclusion of the basepoint of a CW complex is a cofibration, meaning if I have a basepointed map between CW complexes $f:X \rightarrow Y$, given a path $\gamma$ in $Y$ starting at the basepoint, I can extend this path to a homotopy $X \times I \rightarrow Y$ that has its 0-slice $f$. If $\gamma$ is a loop, you can define an action of $\gamma$ on $[X,Y]_*$ by taking $f$ to the 1-slice of the homotopy .

To read more about this and see that it is well defined, read Hatcher's section 4.A.

Now I assume $n>2$, I assume you can do it in the lower cases.

To do this calculation, simply lift the homotopy to $S^{n-1}$ as it is the universal cover, you can see that the effect of this equivalence relation is identifying a map into $S^{n-1}$ with its antipode, so if $n-1$ is even you are quotienting out by the action of postcomposition by a $-1$ degree map, and if $n-1$ is odd you are doing nothing.

Note, postcomposing by $-1$ is not the same as taking the negative. This is only true after suspending. An example, the Hopf fibration is homotopic to itself postcomposed by $-1$, but it is not order 2. However, its suspension is order 2. I'm not sure what is known in general about this postcomposition.

However, for the case of $[S^n, \mathbb{R}P^{n-1}]$ we can explicitly deal with this. If $n>3$ the nth homotopy group of $S^{n-1}$ has stabilized to $\mathbb{Z}/2$, so we can indeed say that postcomposition by a degree -1 map is negation (which is the identity). So $[S^n,\mathbb{R}P^{n-1}]=\mathbb{Z}/2$ if $n>3$. If $n=3$, then we have $\pi_3(S^2)$ generated by the Hopf fibration which I mentioned is not affected by the postcomposition, so $[S^n,\mathbb{R}P^{n-1}]=\mathbb{Z}$.

So it is actually the case that the unbased maps are the same as the based maps for this specific choice of sphere and projective space, but it won't be true in general.

Connor Malin
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