Suppose we arrange, in 3-dimensional space, 8 identical solid cubes in space so they form a square-shaped ring (using a 3x3 arrangement of squares except for the one in the middle).
Its surface will be topologically a torus, composed of a total of 32 congruent squares, any two of which intersect each other along an entire common edge or a common vertex, or not at all.
If we now remove one of the 32 congruent squares, the 31 that remain will form a torus with a hole punched through it, and such that any two of the squares that intersect do so only along an entire common edge or a common vertex.
Puzzle:
What is the smallest number of congruent squares that you can arrange in 3-dimensional space so that they form a topological torus with a hole punched through it, and such that any two squares that intersect each other do so along an entire common edge or along a common vertex?