Let us consider the closed disk $\overline{\mathbb{B}(0,1)} \subsetneqq \mathbb{R}^2$. Let moreover $\mathbb{RP}^2:=\overline{\mathbb{B}(0,1)}/\sim$, where the equivalence relation $\sim$ identifies two antipodal points. Given $f:(x,y) \in \overline{\mathbb{B}(0,1)} \to (-x,-y) \in \overline{\mathbb{B}(0,1)}$, it induces a continuous map $\tilde{f}: \mathbb{RP}^2 \to \mathbb{RP}^2$. In fact, denoted by $\pi$ the classical projection onto the quotient, we may define $\tilde{f}([(x,y)])=[f(x,y)]$. As $\pi \circ \tilde{f}=f \circ \pi$, $\tilde{f}$ has to be continuous. At this point, I consider the homomorphism induced on the fundamental group of $\mathbb{RP}^2$, i.e. $\tilde{f}_*: \pi_1(\mathbb{RP}^2)=\mathbb{Z}/2\mathbb{Z} \to \pi_1(\mathbb{RP}^2)=\mathbb{Z}/2\mathbb{Z}$? What is such a homomorphism?

  • $\begingroup$ Note that $\mathbb Z/2$ has quite few maps to itself. Your map $\tilde{f}_*$ is either the identity or zero. Can it be zero? $\endgroup$ – Pedro Tamaroff Sep 19 '18 at 9:52
  • $\begingroup$ $\tilde{f}$ is the identity $\endgroup$ – Max Sep 19 '18 at 9:54
  • $\begingroup$ Could you write an answer where you prove in detail that $\tilde{f}$ is continuous and that $\tilde{f}_*$ is the identity, please? $\endgroup$ – TheWanderer Sep 19 '18 at 10:22

The map $f$ is a homeomorphism. The maps $\tilde{f}$ and $\tilde{f}_*$ are then induced by functoriality, hence $\tilde{f}_*$ is an isomorphism. The only self isomorphism of $\mathbb{Z} / \mathbb{2Z}$ is the identity, so $\tilde{f}_*$ has to be the identity.

Edit: I fill in some details as requested.

The map $\tilde{f}$ is induced by the universal property of the quotient applied to $f$ followed by the quotient map $\pi$. It is an isomorphism because the quotient is unique and $\pi\circ f=\pi$ ($\tilde{f}$ is in fact the identity).

The map $\tilde{f}_*$ is then induced by the functor $\pi_1$. This being a functor means that it preserves isomorphisms (because it preserves composition and identities). Hence $\tilde{f}_*$ is a self isomorphism of $\mathbb{Z}/2\mathbb{Z}$, which must be the identity (and in fact if you believed before that $\tilde{f}$ is the identity then you don't need to argue that much, the functor just sends identity to identity).

  • $\begingroup$ is the projection map $\pi$ open in this case? How is $\tilde{f}_*$ induced? Can you edit your answer in a more formal way please? $\endgroup$ – TheWanderer Sep 19 '18 at 13:48

A nonalgebraic point of view is that $ \pi \circ f$ induces a map $\widetilde{\pi f}: \mathbb RP^2 \to \mathbb RP^2$ and this map is clearly identity since $\pi \circ f=\pi$.


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