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?
On
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.
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 $E - V + 1 = 5$.