Opening a cube does not change its homotopy type Consider the (skeleton of a) cube. I want to compute the fundamental group.
Now I know that if you "open the cube" - like you would open a box - or in another pov you splat it onto a plane, what you get is something which is homotopy equivalent to a bouquet of five circles, hence has the free product of five copies of $\mathbb{Z}$ as fundamental group.
My question is

Why "opening" the cube does not change its homotopy type?

 A: As far as I can tell this is a coincidence, sort of. You can write down a homotopy that "opens" the cube by "unzipping" four edges - it would be a lot easier to draw diagrams describing this than to describe it verbally but I don't have an easy way to draw diagrams right now.
In any case you don't need this. The general result is that a connected graph with $V$ vertices and $E$ edges is homotopy equivalent to a bouquet of $E - V + 1$ circles (the standard proof is by contracting any spanning tree to a point; you can also just repeatedly contract edges until you can't anymore). For the $1$-skeleton of the cube, $E = 12, V = 8$, so $E - V + 1 = 5$.
A: It doesn't change the homotopy type because it doesn't even change the homeomorphism type. Let $X$ be the skeleton of a cube and $Y$ be the planar graph with $6$ vertices and $12$ edges you have in mind. Define a bijection $X\to Y$ sending vertices of $X$ to vertices of $Y$ and corresponding edges of $X$ to edges of $Y$ (like a central projection from a point above the cube would). This map is a homeomorphism and in particular a homotopy equivalence.
Now you can contract edges in $Y$ to obtain homotopy equivalence with a bouquet of circles.
