Fundamental group of space form by gluing boundary of Mobius band to $S^1 \vee S^1$

Question

Let $X$ be the space formed by gluing the boundary of the Mobius band around $S^1\vee S^1$ along the path $a \cdot b$ (see figure.) I would like to use Van Kampen's theorem to find a presentation for the fundamental group of $X$.

Using $U$ and $V$ as shown in the picture, we see that $V \simeq \{point\}$, $U \simeq S^1\vee S^1$, $U \cap V \simeq S^1$, so

$\pi_1(V) \cong \{e\},\; \pi_1(U) \cong \Bbb Z\ast\Bbb Z=<a,b>, \;\pi_1(U\cap V)=<c>$

Including a generator of $\pi_1(U \cap V)$ into $V$ adds no knew relations since $V$ is simply connected. Is it correct that including into $U$ adds the relation $c=ab$, since travelling around $c$ is the same as travelling around $a$ then $b$? This would imply by Van Kampen's theorem that

$\pi(X,x)\cong <a,b,c \ | \ c = ab>$

for any $x$ in $U \cap V$.

Thanks in advance!

Answer 1

First of all we have to understand all the occuring inclusion maps on $\pi_1$. As mentioned in my comment $V$ is a circle and therefore we have to consider the following to inclusions from $\pi_1(U\cap V)$ in to $\pi_1(U)$ and $\pi_1(V)$ respectively.

It is easy to see that the inclusion $\pi_1(U\cap V) \to \pi_1(U)$ is the map sending a generator to $a\cdot b$ and the other inclusion $\pi_1(U\cap V) \to \pi_1(V)$ is given by multiplication by $2$. From this we conclude that $\pi_1(X)=\pi_1(U)\ast \pi_1(V)/\sim$, where $\sim$ identifies two times the generator of $\pi_1(V)$ with $a\cdot b$ in $\pi_1(V)$