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Does every continuous function $f: \mathbb R P^2 \vee \mathbb R P^2 \to \mathbb R P^2 \vee \mathbb R P^2$ have a fixed point?

I don't really have a good feeling as to whether or not this is true. My first thought was to try and get a map $f':\mathbb R P^2 \to \mathbb RP^2$, which was easier to show has a fixed point, but I can't seem to restrict $f$ in an appropriate way.

If $X = \mathbb R P^2 \vee \mathbb R P^2$, then any map $f:X \to X$ induces a map $\varphi:\tilde X \to \tilde X$, where $\tilde X$ is the universal cover. I believe $\tilde X$ is a (infinite) wedge product of spheres. If I can show that $\varphi:\tilde X \to \tilde X$ has a fixed point, then it should follow that $f$ does, too, by doing some commutative diagram chasing. However, I can't prove that final fact (if it is true, at all).

I'm doing some algebraic topology studying on my own before I take the class next semester. This is coming from an old qual, so hints would be great. Thanks!

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Let $x$ denote the intersection point between the two copies of $\mathbb{R} P^2$. If $f(x) = x$ we are of course done, so assume $x$ is sent to the "interior" of one of the two copies of $\mathbb{R}P^2$. Suppose it is the first copy. Now restrict the domain of $f$ to the first copy, and compose $f$ with the projection to the first copy, to get a map $\mathbb{R}P^2 \to \mathbb{R}P^2$. If this map has a fixed point, then so did $f$.

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  • $\begingroup$ Thanks! This was kind of the idea I had, but I couldn't think of restricting the image, and how the intersection point mattered. $\endgroup$
    – Ben Tighe
    Dec 12 '17 at 4:12

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