# Use Seifert-van Kampen to compute the fundamental group

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?

## 1 Answer

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 $$\pi_1$$