Fundamental group of the surface of a cube with interior of all edges removed

Question

Find the fundamental group of the surface of a cube with interior of all edges removed (i.e. the space which consists of the vertices and interior of the faces of the cube.

Can I deformation retract the interior of each face into a cross meeting in the middle of the face made up of two diagonal lines which connects the each vertex to the vertex opposite of it. I can deform this into a sphere which has a lattice on it with 12 different loops. So the fundamental group is $\mathbb{Z}^{12}$?

Thanks!

Answer 1

The answer by rewritten is correct, although your intuition is also good.

The specific deformation that you consider will give you the correct fundamental group, but as rewritten says, the loops do not commute so you will get $\mathbb{Z} * \cdots * \mathbb{Z}$ for some number of copies of $\mathbb{Z}$. What is that number?

Well, if you think of the resulting object as a sphere less some points, you can see pretty easily that if $X_d = S^2 \setminus \{p_1, \ldots, p_d\}$, then:

• $\pi_1(X_1) = 0$, since this is just a sphere less a single point i.e. a disk.
• $\pi_1(X_2) = \mathbb{Z}$, since this is just a disk less a point (i.e. an annulus).
• $\pi_1(X_3) = \mathbb{Z} * \mathbb{Z}$ since this is just an annulus less a point, which deformation retracts onto a wedge of two circles.

If you continue this on, you see that $\pi_1(X_d) = \underbrace{\mathbb{Z} * \cdots * \mathbb{Z}}_{d-1}$.

Answer 2

It is probably the free product $\mathbb{Z}^{\ast11} = \mathbb{Z}\ast\dots\ast\mathbb{Z}$, as the loops do not commute.

An easier way is that you can "continuously" (not exactly an equivalence but in this case it doesn't matter) deform your set into a cube with $12$ points removed, which is topologically equivalent to a disk with $11$ points removed.