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?
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 $\pi_1$