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I'm preparing for an exam by looking at an old exam archive. I encountered this problem which I believe should be solvable using only material from James Munkres' Topology or possibly John Lee's Introduction to Smooth Manifolds as that is how these exams are designed:

"Let $\mathbb{R}\mathbb{P}^n$ denote the $n$-dimensional projective space where $n$ is an odd integer. Suppose that a continuous map $f:\mathbb{R}\mathbb{P}^n \to \mathbb{R}\mathbb{P}^n$ induces the trivial homomorphism

$$f_*:\pi_1(\mathbb{R}\mathbb{P}^n,x_0) \to \pi_1(\mathbb{R}\mathbb{P}^n,f(x_0)).$$

Show that $f$ has a fixed point."

I was thinking that perhaps $n$ being odd is important mainly because $\mathbb{R}\mathbb{P}^n$ is orientable in those cases. My attempt so far: lift $f$ to a map $\hat{f}:\mathbb{R}\mathbb{P}^n \to S^n$ which is possible since $f$ is trivial and so $f_*(\pi_1(\mathbb{R}\mathbb{P}^n)) \subset p_*(\pi_1(S^n))$ where $p:S^n \to \mathbb{R}\mathbb{P}^n$ is the usual quotient map. From there, I was considering $\hat{f} \circ p: S^n \to S^n$ and trying to say something about the degree of the map or using something like Brouwer Fixed Point Theorem (first extend $f$ because it is nullhomotopic).

So far, some friends have suggested using suspensions (as in Hatcher) or the Lefschetz fixed point theorem but I think these are too advanced and wouldn't score points on the exam. Any hints would be a huge help; thank you.

Edit: I've written up a possible solution below; let me know what you think.

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  • $\begingroup$ Can you sketch a solution using Lefschetz fixed point theorem? Would you be using the cohomology ring of $\mathbf RP^n$ in $\mathbb Z$ coefficients? $\endgroup$ May 29 '18 at 2:30
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I think I have a solution (to my own question) which uses sufficiently elementary theory. I owe some thanks to friends for it.

Suppose that $f$ has no fixed point. Since $f_*$ is trivial, as I said in the question details, it lifts to a map $\hat{f}:\mathbb{R}\mathbb{P}^n \to S^n$. If we let $g = \hat{f} \circ p$, then $g: S^n \to S^n$ is a map such that $p \circ g = f \circ p$.

Claim: If $f$ has no fixed point, neither does $g$ on the $n$-sphere.

  • For, if $g(x)=x$, then $p \circ g(x) = [x] = f \circ p(x) = f[x] = [x]$ (equivalence class $x \sim -x$). Then $f$ has a fixed point which is a contradiction.

Claim: $g$ is homotopic to the antipodal map $\alpha(x) = -x$.

  • Let $H:S^n \times I \to S^n$ be defined by

$$H(x,t) = \frac{(1-t)g(x) + t \alpha(x)}{\|(1-t)g(x) + t \alpha(x)\|}.$$

The denominator is never zero because for it to be zero, we must have that $(1-t)g(x) -tx = 0 \Rightarrow g(x) = \frac{t}{1-t}x$. Since $\|g(x)\| = 1 = \|x\|$, we need $t = 1/2$. But then $g(x) = x$ and $g$ has no fixed points; a contradiction. Thus, $H$ is a homotopy for $g$ to $\alpha$.

Claim: When $n$ is odd, $\mathbb{R}\mathbb{P}^n$ is orientable and the antipodal map on $S^n$ is homotopic to the identity map.

  • I won't prove the first claim; I think it's a standard fact. As for the second, it is proved in the usual proof for the Hairy Ball Theorem (See Problem 16-6 in Lee's Smooth Manifolds). It relies on the fact that when $n$ is odd, there exists a nonvanishing vector field $V$ on $S^n$. We can make a homotopy from $\alpha$ to $V$ and from $V$ to the identity map.

Thus, if $f$ has no fixed points, $g$ is homotopic to the identity. Now, the degree of a map is homotopy invariant and since $\text{id}$ has degree 1, $\text{deg} \,g= 1$. Now, since $f \circ p = p \circ g$, $\text{deg}(f \circ p) = (\text{deg} \, f)(\text{deg} \, p) = (\text{deg} \, p)(\text{deg} \, g) = \text{deg}(p \circ g)$. So $\text{deg} \, f = \text{deg} \, g = 1$. By the way, degree is well-defined here since $\mathbb{R}\mathbb{P}^n, S^n$ are compact, orientable manifolds of the same dimension.

Claim: If $f_* =0$, then $\text{deg} \, f$ is even.

  • We consider the map $f^*: H^1(\mathbb{R}\mathbb{P}^n, \mathbb{Z}_2) \to H^1(\mathbb{R}\mathbb{P}^n, \mathbb{Z}_2)$. Now, by the Hurewicz theorem, $H_1(X) = \text{Ab}(\pi_1(X))$. Since $\pi_1(\mathbb{R}\mathbb{P}^n) = \mathbb{Z}_2$ is already abelian, then $H^1(\mathbb{R}\mathbb{P}^n, \mathbb{Z}_2) = \text{Hom}(H_1(\mathbb{R}\mathbb{P}^n, \mathbb{Z}_2), \mathbb{Z}_2) = \text{Hom}(\mathbb{Z}_2,\mathbb{Z}_2) = \mathbb{Z}_2$. So $f^*: \mathbb{Z}_2 \to \mathbb{Z}_2$. I want to say from this that $f^* = 0$ because $f_* = 0$.

If this is correct, then the conclusion is: $\text{deg} \, f = 1$ but should also be even. This is a contradiction and thus, $f$ has a fixed point.

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    $\begingroup$ I'm not sure about your last claim. But you don't need it. You've proven that $\deg f = 1$. Now, argue that $f^\ast:H^1(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})\rightarrow H^1(\mathbb{R}P^n; \mathbb{Z}/2\mathbb{Z})$ is the $0$ map because $f_\ast$ is the $0$ map. Use the ring structure of $H^\ast(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})$ to conclude the mod $2$ degree of $f$ is $0$. In other words, the degree of $f$ must be even, contradicting the fact that $\deg f = 1$. $\endgroup$ May 30 '18 at 3:25
  • $\begingroup$ @JasonDeVito I'm having some trouble arguing $f^* = 0$ and also that the degree of $f$ is even by looking at the ring structure. $\endgroup$
    – inkievoyd
    May 30 '18 at 15:35
  • $\begingroup$ I've added an answer addressing your comment. Let me know if I can further help! $\endgroup$ May 30 '18 at 16:01
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(This meant to answer a question in the comments)

Let $X:=\mathbb{R}P^n$ and let $G:=\mathbb{Z}/2\mathbb{Z}$, just to save a bit of typing.

Lemma 1: If $f_\ast:\pi_1(X)\rightarrow \pi_1(X)$ is the $0$ map, then so is $f_\ast:H_1(X)\rightarrow H_1(X)$.

Proof: $H_1$ is the abelianization of $\pi_1$, which is already abelian. Further, the abelianization is natural, so $f_\ast$ is $0$.

Lemma 2: If $f_\ast:H_1(X)\rightarrow H_1(X)$ is the $0$ map, then so is $f^\ast H^1(X;G)\rightarrow H^1(X;G)$.

Proof. From universal coefficients, we know that $H^1(X;G) \cong Hom(H_1(X), G)\oplus Ext(H_0(X),G)$. Since $H_0(X)$ is free, $Ext(H_0(X),G) = 0$. The the isomorphism $H^1(X;G)\rightarrow Hom(H_1(X),G)$ is natural.

Now, $f_\ast:H_1(X)\rightarrow H_1(X)$ induces a map $\hat{f}$ on $Hom(H_1(x),G)$ given by $\hat{f}(g) = g\circ f_\ast$. If $g\in Hom(H_1(X), G)$ is the unique nontrival map sending $1\in G$ to $1\in G$, then $\hat{f}(g)(1) = g(f_\ast(1)) = g(0) = 0$, so $\hat{f}(g) = 0$. It follows that $\hat{f}:Hom(H_1(X), G)\rightarrow Hom(H_1(X),G)$ is the $0$ map.

By naturality of the isomorphism $H^1(X;G)\rightarrow Hom(H_1(X),G)$, $f^\ast$ must be the $0$-map as well.

Lemma 3: Suppose $f:X\rightarrow X$ induces the $0$ map on $H^1(X;G)$. Then $f$ induces the map on $H^i(X;G)$ for any positive $i$.

Proof:

Recall $H^\ast(X;G)\cong G[x]/x^{n+1}$, where $x\in H^1(X;G)$.

By assumption, $f^\ast(x) = 0$. Then $f^\ast(x^k) = f^\ast(x)^k = 0^k = 0$, so $f^\ast$ is the $0$ map on anything $x$ generates. But $x$ generates all positive degree cohomology.

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