Fundamental Group of Surface Formed by Gluing Three Mobius Strips along their Boundaries Call the above surface $X$. I am attempting to use Seifert van Kampen to compute the fundamental group $\pi_1(X,x)$.
To start, if we call the three Mobius strips $M_1,M_2$, and $M_3$, Seifert van Kampen tells us that if $Y$ is the surface given by gluing $M_1$ to $M_2$ along their boundaries, then $\pi_1(Y,x) = \langle a,b | \ a^2 = b^2 \rangle$.  I have read that this is a presentation of $\pi_1(K,k)$, where $K$ is the Klein bottle.
The problem is that I'm not really sure where to go from here aside from using Seifert van Kampen on $Y$ and $M_3$ glued along their boundaries.  The real issue I am having is in finding the relations in the amalgamation.  To be clear, if $U$ is a fattened neighborhood of $Y$ and $V$ is a fattened neighborhood of $M_3$, then I know that $U$ and $V$ deformation retract to $Y$ and $M_3$, respectively, while $U \cap V$ deformation retracts to the common boundary of $Y$ and $M_3$, which is just a circle $S^1$.  Then all of $U,V,$ and $U \cap V$ are path-connected so Seifert van Kampen applies.  Because $U$ deformation retracts to $Y$ and $V$ deformation retracts to $M_3$, we have that
\begin{equation*}
\begin{split}
\pi_1(U,x) &= \pi_1(Y,x) &= \langle a,b| \ a^2 = b^2 \rangle \\
\pi_1(V,x) &= \pi_1(M_3,x) &= \langle c | \ \rangle \\
\pi_1(U \cap V,x) &= \pi_1(S^1,x) &= \langle g| \ \rangle.
\end{split}
\end{equation*}
Now, if $i: U \cap V \to U$ and $j : U \cap V \to V$ are the standard inclusion maps, then we want $i_*(g)$ and $j_*(g)$ where $i_*$ and $j_*$ are the induced homomorphisms of fundamental groups.  I know that $j_*(g) = c^2$.  However, I am not quite sure what $i_*(g)$ should be and why it should be that.
Of course after knowing $i_*(g)$, I can explicitly have 
$$\pi_1(X,x) = \pi_1(U,x)*_{\pi_1(U \cap V,x)} \pi_1(V,x) = \langle a,b,c | \ a^2 = b^2, \ i_*(g)j_*(g)^{-1} = id \rangle.$$
 A: First off, this is not a surface, just a $2$-complex. Near the boundaries you glue all three of them by, it looks like three pages glued by one specific edge each. So there it's not locally homeomorphic to $\Bbb R^2$.
For the fundamental group, here's a perspective that might help: think of the moebius strips as $\Bbb{RP^2} - D$ where $D$ is a small disk on some chart on the real projective plane. You glue three copies these along $\partial D$. Call $a, b, c$ to be the generators of $\pi_1(\Bbb{RP}^2 - D) \cong \Bbb Z$. $\pi_1(X)$ is then just a quotient of $\langle a, b, c \rangle$ by some relator(s). 
If you take the loop $\partial D$, and slightly push it to make it go inside one of the three copies of $\Bbb{RP}^2 - D$, you see its homotopy class is $a^2$, $b^2$ and $c^2$ (as those are the homotopy class of the boundary of each of the moebius strips). So the relator is precisely $a^2 = b^2 = c^2$, making the fundamental group to be $\langle a, b, c |a^2 = b^2 = c^2 \rangle$.
To do this rigorously, do it like you said: think of $X$ as Klein bottle with a moebius strip $M$ with $\partial M$ glued along a "longitude" of the bottle. This longitude has homotopy class $a^2$ ($ = b^2$) according to your presentation of $\pi_1(K)$, so by van Kampen $\pi_1(X) = \pi_1(K) * \pi_1(M)/\langle a^2 = c^2\rangle$, which is exactly the presentation above. 
