Page 171 - A Course in Linear Algebra with Applications

6.1: Functions Defined on Sets 155

F and G are to be considered equal - in symbols F = G - if
they have the same domain and codomain and if F(x) = G(x)
for all x.

Theorem 6.1.1
Let F : X —>Y, G : U —> V and H : R —> S be functions such
that \m{H) is contained in U and Im(G) is contained in X.
Then F o (G o H) = (F o G) o H.

Proof
First observe that the various composites mentioned in the
formula make sense: this is because of the assumptions about
Im(H) and lm(G). Let x be an element of X. Then, by the
definition of a composite,

F o (G o H){x) = F((G o H)(x)) = F(G{H(x))).

In a similar manner we find that (FoG) oH(x) is also equal to
this element. Therefore Fo(GoH) = (FoG)oH, as claimed.

Another basic result asserts that a function is unchanged
when it is composed with an identity function.

Theorem 6.1.2

If F : X —>Y is any function, then F o l x = F — y o F.
l
The very easy proof is left to the reader as an exercise.

Inverses of functions
Suppose that F : X —> Y is a function. An inverse of
F is a function of the form G : Y —> X such that FoG and
G o F are the identity functions on Y and on X respectively,
that is,
F{G{y)) = y and G(F(x)) = x
for all s i n l and y in Y. A function which has an inverse is
said to be invertible.