Fundamental group of the the union of three planes and $S^2$

Question

Let $X,Y$and $Z$ be the $yz,xz,xy$ planes respectively and let $B=X\cup Y\cup Z\cup S^2$ where $S^2$ be a space with the subspace topology is the unit 2-sphere. I want to find the fundamental group of $B$. I have a idea but no rigorous proof to give. The three planes creates 8 bounded empty space inside $S^2$. If everything outside can be retracted to the boundary of the 2-sphere we have a sphere that bounds 8 hollow spaces that lie within the sphere. This I think can be transformed to a bouquet of 8 petals (hollow). Maybe how I picture this is wrong. Any ideas? Any tips as to how I can use SVK theorem will also be useful. Thanks !

Answer 1

Indeed, everything outside the sphere in $B$ can be retracted to the sphere, and you're right in that you can do better and retract $X\cup Y\cup Z$ all to a single point to obtain the bouquet of 8 petals you describe. Another name for the bouquet is the wedge sum of 8 copies of $S^2$.

Now one of the consequences of TVK is a general theorem $\pi_1$ converts wedge sum to free products. It's a basic result proved in Harcher. Then you obtain the fundamental group of $B$ is the free product of trivial groups and so is trivial.