Is every map $\widetilde{f}:\mathbb{RP}^m \rightarrow \mathbb{RP}^n$ induced from a map $f:\mathbb{S}^m \rightarrow \mathbb{S}^n$?.

In other words, let $\mathcal{S}$ be the category of spheres $\mathbb{S}^n$ with $n \in \mathbb{N}_0$ and morphisms given by maps $f: \mathbb{S}^m \rightarrow \mathbb{S}^n$ satisfying $f(x) = f(-x)$. Let $\mathcal{P}$ be the full subcategory of $\mathsf{Top}$ on the projective spaces $\mathbb{RP}^m$. There is a functor $\mathsf{Proj}: \mathcal{S} \rightarrow \mathcal{P}$ sending $\mathbb{S}^m$ to $\mathbb{RP}^m$. Is this functor full?

I tried to construct a section $\mathbb{RP}^m \rightarrow \mathbb{S}^m$, but it did not work out and in retrospect I believe this approach to be bound to fail, since $\mathbb{S}^m$ is orientable, while $\mathbb{RP}^m$ is not. Further I tried to use the universal property of the construction of $\mathbb{RP}^m$, but to no avail. I did not find a counterexample yet.

The question arose when writing down the theorem that every map $f: \mathbb{S}^{2m} \rightarrow \mathbb{S}^{2m}$ admits a point with $f(x) = \pm x$. I was wondering, whether it implies that "every map $g:\mathbb{RP}^{2m} \rightarrow \mathbb{RP}^{2m}$ admits a fixed point".


Yes, this follows from the theory of covering spaces but we have to divide this into a few cases, first assume $m>1$, no restriction on $n$. Define the projection map:

$$p':S^n \rightarrow \mathbb{RP}^n$$

We have a map $g$ given by the composition

$$S^m \xrightarrow p \mathbb{RP}^m \xrightarrow{\tilde{f}} \mathbb{RP}^n$$

such that $g_*(\pi_1(S^m)) \subset p'_*(\pi_1(S^n))$, (Since $\pi_1(S^m)=0$ if $m>1$)

Which means that because $p'$ is a covering map $g$ has to have a lift along $p'$ which is equivalent to saying that there exists a map $f:S^m \rightarrow S^n$ such that $p' \circ f = g =\tilde{f}\circ p$. Which is identical to saying that $\tilde{f}$ is induced by $f$.

Now when $m=1$ and $n>1$, $\mathbb{RP}^m = S^1$, $\tilde{f}$ will be induced by an $f$ since $p_*$ is multiplication by two on homotopy groups, identifying $\pi_1(S^1)=\pi_1(\mathbb{RP}^1)$ and thus every element in the image of $g_*$ will be a multiple of two, so zero since $\pi_1(\mathbb{RP}^n)=\mathbb Z_2$ for $n>1$ which means that $g_*(\pi_1(S^m)) \subset p'_*(\pi_1(S^n))$ since every element in the image of $g$ is zero. And a lift exists by the previous argument.

Now if $m=1$ and $n=1$ we can just take $f=\tilde{f}$ if we fix a homeomorphism $S^1 \rightarrow \mathbb{RP}^1$ and apply it to both $\mathbb{RP}^n$ and $\mathbb{RP}^m$.

  • 1
    $\begingroup$ Thank you very much! I will have to refresh my understanding of covering spaces and fundamental groups, but it in my mind this is a very nice result. $\endgroup$
    – PrudiiArca
    Feb 18 '20 at 23:05

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