let $f: D_2 \to D_2$ such that $f (x, y) = (- x, -y)$. Show that $f$ induces a continuous map $f ^{'}: \mathbb{RP}^2 \to \mathbb{RP}^2$. Then determine the induced homomorphism on the fundamental group $f ^{'*}: \pi_1(\mathbb{RP}^2, x_1) \to \pi_1(\mathbb{RP}^2, x_1)$, where $x_1$ belongs to $\mathbb{RP}^2$. I tried to apply path lifting theorem but I don't know how to do it.
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do you mean $S^2\to S^2$ ? – Maxime Ramzi Mar 16 '19 at 20:15
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no sorry D2 to D2 – Giusy Mar 16 '19 at 20:21
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How are you defining projective space? – Exit path Mar 16 '19 at 20:33
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The induced map $\mathbb{R}P^2 \rightarrow \mathbb{R}P^2$ is obtained by looking at the composition $D^2 \rightarrow D^2 \rightarrow \mathbb{R}P^2$ where the last map is the quotient of opposite points on the boundary. Then if we compare where $x,-x$ are mapped under the composition, we see that if $x$ is on the boundary of $D^2$ then $x,-x$ get mapped to the same point. Properties of the quotient then tells us we can define a continuous map $\mathbb{R}P^2 \rightarrow \mathbb{R}P^2$ by lifting a point in $\mathbb{R}P^2$ up to the disk and mapping it under the composition above.
Now draw out your diagram of $\mathbb{R}P^2$ (I prefer the second image on this page Fundamental Polygon of Real Projective Plane), we know that one of these semicircles generates the fundamental group, so all we need to know is how it acts on this loop. Well there is an obvious lift up to $D^2$ by just erasing the arrows, then you map it to it's negative, then you do the quotient. So we lifted a point to its fiber under the quotient map, negated it which still kept it in the fiber, then did the quotient map. This means that on this loop it acts like the identity. So the map on fundamental groups is the identity.
Connor Malin
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