fundamental group of projective plane with two indentified points

Question

I tried to calculate fundamental group of projective plan with two indentified points. I found that this is the free product between fundamental group of projective plan and circle, i.e. $\mathbb Z_2 \ast \mathbb Z$. I would know that if it is correct.

Answer 1

It is, and I don't know how you proved it but here's a sketch of a proof : Let $Y$ be your quotient of the projective plane.

Let $X$ denote the projective plane union a path $P$, where the endpoints of the path identified to the two points you're identifying, then $X$ is a CW-complex, and the inclusion $P\to X$ is the inclusion of a sub-CW-complex, thus it is a cofibration, and $X/P$ is homotopy equivalent to $X$. Note that $X/P\cong Y$, so that $X$ and $Y$ are homotopy equivalent.

Now let $L$ be a path between those two points in $\mathbb{R}P^2$ (or $n$). Then for the same reason, $X/L$ is homotopy equivalent to $X$, therefore to $Y$.

However, $\mathbb{R}P^2/L$ is homotopy equivalent to $\mathbb{R}P^2$ for the same reason, and $X/L$ is homeomorphic to $\mathbb{R}P^2/L \lor S^1$, thus homotopy equivalent to $\mathbb{R}P^2\lor S^1$.

Therefore $Y$ is homotopy equivalent $\mathbb{R}P^2\lor S^1$, and then you can use the Van Kampen theorem to conclude.