Fundamental group of $S^{1}$ unioned with its two diameters Is my solution correct?
Call $X$ the riflescope space (I made this name up). I let $p$ be the point of intersection of the two diameters, and $q$ be the right point of intersection of the horizontal diameter with the circle. I let $A=X-{p}$ and $B=X-q$. Then $A \cap B = X-p-q$. $A$ and $B$ are open path connected with $X = A \cup B$. Furthermore $A \cap B$ is simply connected since it deformation retracts to a point. $A$ deformation retracts onto $S^{1}$ so has fundamental group $\mathbb{Z}$ and $B$ deformation retracts onto a space homotopy equivalent to the theta space which I know has fundamental group $\mathbb{Z}*\mathbb{Z}$. I conclude by van Kampen that $\pi_{1}(X) \simeq \mathbb{Z}*\mathbb{Z}*\mathbb{Z}$.

Okay so my original solution isn't correct. I have a feeling that it's the product of 4 copies of $\mathbb{Z}$ but I'm not sure how to decompose $X$ yet. 


If your space looks like $\bigoplus$ then it is homotopy equivalent to a wedge of four circles (collapse the diameters to a point) so Van Kampen gives its fundamental group is the free group on four generators (not three). 
