How to show that $f_* (\sigma)=\sigma$ where $f$ is mapping between projective spaces $\mathbb{R}\text{P}^3$

Question

Suppose that $f:\mathbb{R}\text{P}^3 \to \mathbb{R}\text{P}^3$ is continuous mapping without fix points and let $\sigma$ be (some) generator of group $H_3(\mathbb{R}\text{P}^3)$. Prove that $f_{*}(\sigma)=\sigma$.

I have this problem today on exam and I failed to solve it. I would really appreciate if someone could show me how to solve problems like this one.

Edit: Just to add that we didn't learn Lefshets theorem, so answer from @Melo.M.M is to advanced for me.

Answer 1

We prove this for any $\mathbb{R}P^n$ with $n > 1$ odd. The first half of the proof works regardless of the parity of $n>1$.

Let $\pi:S^n\rightarrow\mathbb{R}P^n$ be the universal covering. Let $f:\mathbb{R}P^n\rightarrow \mathbb{R}P^n$ be any continuous function. Then, by the lifting criteria, there is a function $g:S^n\rightarrow S^n$ with $\pi \circ g = f\circ \pi$.

Now, assume $f$ has no fixed points. I claim that $g$ has no fixed points. For, if $g(x) = x$. Then $\pi(x) = \pi(g(x)) = f(\pi(x))$, so $\pi(x)$ is fixed by $f$.

Now, we have the following lemma:

Lemma Suppose $g:S^n\rightarrow S^n$ has no fixed points. Then $g$ is homotopic to the antipodal map $a(x) = -x$.

Proof: Define $F:S^n\times I\rightarrow S^n$ by $$F(x,t) = \dfrac{tg(x) + (t-1)x}{|tg(x) + (t-1)x|}.$$

Note that this is well defined: If $tg(x) + (t-1)x = 0$, then $|tg(x)| = |(t-1)||x|$ so $t = \frac{1}{2}$. Plugging this back in gives $g(x) -x = 0$, i.e., $g(x) = x$, contrary to the fact that $g$ has no fixed points.

Now simply note that $F(x,1) = g(x)$ and $F(x,0) = -x$, so $F$ is a homotopy beteen $g$ and the antipodal map. $\square$

The rest of the proof needs the assumption that $n$ is odd. Recall the well known fact that when $n$ is odd, the antipodal map, and therefore $g$, is homotopic to the identity map. (I can sketch a proof if you need).

To finish off the proof, we have $\pi_\ast:\mathbb{Z} \cong H_n(S^n)\rightarrow H_n(\mathbb{R}P^n)\cong \mathbb{Z}$ is given, up to sign, as multiplication by $2$. So, given $\sigma \in H_n(\mathbb{R}P^n)$ there is a generator $\nu\in H_n(S^n)$ with $\pi_\ast(\nu) = 2\sigma$.

Then we have $2f_\ast(\sigma) = f_\ast(2\sigma) = f_\ast\pi_\ast(\nu) = \pi_\ast g_\ast( \nu) = \pi_\ast(\nu) = 2\sigma$.

Since $2$ is cancellable in $H_n(\mathbb{R}P^n) \cong \mathbb{Z}$, this implies $f_\ast(\sigma) = \sigma$ as desired.

Answer 2

I think so.. We suposse that $f_{*}\neq id$, then $$L(f)=\sum_{k\geq 0} tr(f_{*}:H_k(\mathbb{R}^3P)\to H_k(\mathbb{R}^3P)) = tr(f_{*}:H_0(\mathbb{R}P^3)\to H_0(\mathbb{R}P^3)) - tr(f_{*}:H_3(\mathbb{R}P^3)\to H_3(\mathbb{R}P^3)) = 1 - tr(f_{*}:H_3(\mathbb{R}^3P)\to H_3(\mathbb{R}P^3)) \neq 0$$ then for Lefshets theorem $f$ has fix point, contracdition...