0
$\begingroup$

I am trying to generalize the problem I ask yesterday Fundamental group of sphere with antipodal points on the equator, i.e. the question is "Compute the fundamental group of the space obtained from two copies of $\mathbb{R} P^{n+1}$ by gluing them along standard copies of $\mathbb{R} P^n$.".

The hard part, in my opinion, is that given the natural inclusion $\mathbb{R}P^n\to \mathbb{R}P^{n+1}$, how do we know the map $\pi_1(\mathbb{R}P^n)\to \pi_1(\mathbb{R}P^{n+1})$?

The answer in the link can be used when $n\geq 2$ because then $\mathbb{R}P^n$ are the same in the sense that they admit $S^n$ as their 2-sheeted universal cover. So what if $n=1$? Then the question is, what do we know about the map $f:\mathbb{Z}\to \mathbb{Z}_2$? Is it true that $f(1)=1$ or $f(1)=0$?

The difference of this case is that $\mathbb{R}P^1$ admits $\mathbb{R}$ as its infinitely-sheeted universal cover. SO we can not use the same argument. On the other hand, it's easier to consider $n\geq 2$ since we only have to deal with the 2-element groups $\mathbb{Z}_2$. In this case, a group $\mathbb{Z}$ appears. So there are a lot of elements to take out here.

THese are my difficulties trying to apply the same argument. This post Fundamental group of 2 copies of $\mathbb{R}P^2$ glued along a common $\mathbb{R}P^1$ uses a CW structure to solve the same problem so I know that $f(1)=1$. But can we do it using Seinfert-van Kampen?

$\endgroup$
2
$\begingroup$

So if I understand correctly you want to know what the natural inclusion $\mathbb RP^1\to \mathbb RP^2$ induces on $\pi_1$ ?

We can still use the same kind of idea : we have a commutative square, where $S^1\to S^2$ is the natural inclusion, and the vertical maps are the natural covering maps :

$$\require{AMScd}\begin{CD}S^1@>>> S^2 \\ @VVV @VVV \\ \mathbb RP^1@>>> \mathbb RP^2\end{CD}$$

Note that $S^1\to \mathbb RP^1$ indues $\mathbb Z\overset{2}\to \mathbb Z$ on $\pi_1$ when you identify them with $\mathbb Z$

Fix a basepoint $1\in S^1$, and call $x$ its image in $\mathbb RP^1$. Now take $f:I\to \mathbb RP^1$ a path that generates the fundamental group of $\mathbb RP^1$ at $x$. If you lift that path to a path $g$ in $S^1$ starting at $1$, you also get a path, with endpoint in $p^{-1}(x)=\{-1,1\}$.

The end point can't be $1$, otherwise it would be a loop and our generator of $\mathbb RP^1$ would be in the image of $\pi_1(S^1)$, which we know it can't be, so it's $S^1$.

So if you push $f$ to $\mathbb RP^2$, a lift to $S^2$ can be given by pushing $g$ to $S^2$. But $g$ in $S^2$ isn't a loop, which must mean that $I\to \mathbb RP^1\to \mathbb RP^2$ isn't nullhomotopic (otherwise it would lift to a loop). But the only non nullhomotopic loop in $\mathbb RP^2$ is the generator of $\pi_1(\mathbb RP^2) = \mathbb Z/2$, so $\mathbb RP^1\to \mathbb RP^2$ induces the natural projection $\mathbb Z\to \mathbb Z/2$ on $\pi_1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.