Section 2  Product of Functions

It is very possible to move the objects in U directly to W in one movement going past but not stopping at V. Intuitively speaking, two movements have been combined into a single movement. Let functions f: U --> V and g: V --> W representing the moving of objects from one warehouse to another. f moves each object u in U to V where it becomes f(u). And g will move that object, and any other object in V, to W, where it becomes g(f(u)).

Let function h: U --> W represent the movement from U directly to W. Then that same object u in U becomes h(u) in W.

But h(u) and g(f(u)) are really the same object in W. . Therefore, h and eventually gf do the same thing for each object in U. This means that the two functions are equal: h = gf. Notice that the second warehouse V must receive all the objects that were in U and that W must receive all objects that were in V.

[2.1] Let f: be a function from a set U and let g be a function from a set V. Then g is conformable to a function f if the range of f is contained in the domain V of g.

Intuitively speaking, if there are three sets, and f is a function from the first set into the second. and g is a function from the second set into the third, then g can pickup any image produced by f and carry it into the third set.

[2.2] Let a function g be a function conformable to a function f. The product gf of f and g is defined by

Notice that f acts first on x then g acts on the image f(x). However, the notation gf is in a sense written backwards, instead of the normal reading of left-to-right. In some parts of modern algebra and in the discussion of permutations the notations ((x)f)g and fg are used to preserve normal reading order. In these discussions that order of writing products of functions is not used. The "backward" notation is used here.

As an example, let A,B,C be sets of integers as follows: