How can I geometrically understand the fundamental group of the 1-skeleton of the unit cube?

Question

Consider the unit cube $I^3$. I'm aware that the 1-skeleton consists of all the edges of the cube and that the fundamental group is $\mathbb{Z} \star \mathbb{Z} \star \mathbb{Z}$, but I'm trying to understand how the cube deforms into the wedge sum of three circles.

Answer 1

Given a connected, finite graph (such as the 1-skeleton of your unit cube), the following operations do not change the homotopy type (left as an exercise):

1. Removing loose ends, i.e. collapsing edges that have a vertex to which no other edge is attached (that is, a vertex of valence one).

2. Collapsing an embedded edge, i.e. collapsing an edge whose two vertices/ends do not coincide, to a point. (small exercise: Why does this mean that the edge is embedded?)

These processes reduce the number of edges and vertices of a graph. In particular, we can reduce any graph of the given type to a graph with only a single vertex, i.e. a bouquet of circles, in this manner.

In your case, it will reduce the graph to a bouquet of 5 circles. This is probably most easily seen by actually drawing the graph and applying the above operations, making intermediate drawings after each step.