# Does every map between real projective spaces come from a map between spheres?

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 $$\mathbb{RP}^m$$.

• 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. Feb 18 '20 at 23:05