Well, first of all, what a cubical complex is, is a set of points and line segments and squares and cubes and so on in R^n where the points have all integer coordinates, and the line segments are between adjacent points, and the squares are little unit squares, and so on, subject to the rule that whenever a complex has an edge, it also has both endpoints of the edge, and whenever it has a square, it has all edges of the square, and whenever it has a cube, it has all faces of the cube, etc. etc.
Let's play a game where you try to get from one cubical complex to another, where the allowed moves are like so:
Any time you see a point that belongs to the complex next to one that doesn't, (which implies that the edge between them also doesn't belong!) you can fill in that edge and that point. Also, you can do the reverse of this move: given any edge, you can delete that edge and one of its endpoints, as long as the result is still a well-formed cubical complex.
Any time you see all-but-one edges of a square belonging to the complex, you can fill in that last edge and also the square bounded by those edges. Also, you can do the reverse of this move.
Any time you see all-but-one faces of a cube belonging to the complex, you can fill in that last face, and also the solid volume bounded by all the faces. Also, you can do the reverse of this move.
Question: does this notion of step-by-step equivalence generate all homotopy equivalences of cubical complexes considered as topological spaces?
If I throw in the ability to add (but not delete) cells to the complex, does this generate all continuous maps from one cubical complex, subject to the right notion of homotopies between maps?
*Did not actually run a fever, I think, but felt mentally loopy enough to call it that.