This is in marked contrast with Cartesian products. Two points <a, b> and <c, d> may be related by the fact that a = c or alternately by b = d. Every point in a fiber bundle is on a definite fiber. In math notation there is a function f: fb → b, from the bundle to what is called the base space. Two points, x & y, are on the same fiber when f(x) = f(y). A point in the base space corresponds to a fiber. The bundle is not like cooked celery, however, the fibers are glued together. There is another topological space called the fiber space F to which all fibers are homeomorphic.

Every point in the base space has a neighborhood such that the Cartesian product of that neighborhood and the fiber space is homeomorphic to the inverse image of the neighborhood. This is the glue. In fewer words: every point in the base space has a neighborhood whose inverse image is like a Cartesian product with the fiber space. These products cannot, however, in general be merged together to form a Cartesian product of the fiber space and the whole base space.

Consider a Möbeus strip. It is sort of like the Cartesian product of a circle and line segment, but not quite. It is, however a fiber bundle with the fibers running the short way. They can’t run the long way because they don’t meet up once around the circle. The fiber space is a line segment and the base space is a circle. There is the group of two elements that flips the fiber space end for end. In general there is a discrete group on the fiber space which can be used to join together these inverse images.

There is some confusion on the status of the group of a fiber bundle. Some definitions allow a group that is bigger than necessary. Some definitions require the group to be given. Let me give a more careful definition of a FB and then say what I think its group should be.

Given is the bundle space B, the base space b and fiber space f. Also given is the projection π which maps B onto b. It is required that for every x∊b there is some neighborhood n of x and a homeomorphic map φ from n×f to π⁻¹(n) such that for all x'∊n and y∊f π(φ(x', y)) = x'. Now consider a loop in b which is the image of a circle. Traversing that loop …
I thought I was going to improve in Steenrod’s definition of the group but I am stuck for now. Roughly it is a group that operates on f homeomorphically to produce the same transformations on f as trips around loops in b, and no other transformations.

The example that originally motivated fiber bundles was the vector space on a manifold. Consider a sphere as the manifold. We will consider the vectors tangent to the sphere at each point on the sphere. These vectors are the bundle of the fiber bundle. Two vectors at the same point are clearly related so the sphere, itself, is the base space. The two dimensional real vector space is the fiber space.