Fundamental group of $4$ spheres pairwise tangent

Question

I intuitively got that the fundamental group of the topological space given by considering $4$ spheres pairwise tangent in $ \mathbb{R}^3$ is $\langle a,b,c\rangle$. My approach was that if you consider the fundamental group of $3$ spheres pairwise tangent you get the same one as the circumference. Applying this to the $4$ triplets of spheres you get when considering this space we intuitively get 4 pairwise tangent circles, that can be deformed to a $1$-skeleton of a tetrahedron, that I know to have fundamental group equal to $\langle a,b,c \rangle$. The problem is, $3$ tangent spheres are homotopy equivalent to the circumference, not homeomorphic and I don't know if the union of subspaces has the same homotopy type as the union of homotopy-equivalent spaces. Another approach I have tried is trying to find an equivalence between this space and a sphere without $4$ points, but I'm having a hard time finding such maps.

Answer 1

Consider the space $X$ to be the 1-skeleton of a tetrahedron, with a sphere glued to every vertex of the skeleton.

It is possible to write down a homotopy equivalence between your space and $X$. By Seifert-Van Kampen you can then show that the fundamental group of $X$ must be $\langle a,b,c\rangle$ as you said.