(
) In this half of the proof, assume that
exists and we must prove that
is one-to-one and onto. To do so, it is convenient for us to use the relation notation, where
is equivalent to
To prove that
is one-to-one, assume that
Alternatively, that means
and
are elements of
. We must show that
Since
and
are in
By the fact that
is a function and
cannot have two images,
and
must be equal, so
is one-to-one.
Next, to prove that
is onto, observe that for
to be a function, it must use all of its domain, namely A. Let
be any element of
Then
has an image under
,
Another way of writing this is
By the definition of the inverse, this is equivalent to
Hence,
is in the range of
Since
was chosen arbitrarily, this shows that the range of
must be all of
(
) Assume
is one-to-one and onto and we are to prove
exists. We leave this half of the proof to the reader.