hello!
I am trying to work this out, not sure where to start even though the theorem appears intuitive! I thought I had proven it but then realised I has assumed the existance of an inverse!
For functions f,g and sets A, B
Given and and
where is the function
Then exists and
To show the exsistance of an inverse requires that f is a bijection? not sure how to derive that from what is given.
Thanks in advance!
No!
If for then (but it should equal x!) and for the same reason .
@matt: By definition, a function has an inverse iff it is a bijection. You can try proving this here by contradiction (twice) - first suppose it is not injective and reach a contradiction, and then suppose it is not surjective and again reach a contradiction.
aha, ok! so, (not via contradiction though),
for
Assume
then
and since
then
implies
since
So f is 1-1
Now to show it is onto,
for
so for each
there is a
so for each
such that
(each element of B is mapped onto)
so f is onto,
so we know f is a bijection
so it has an inverse!
now... how do we show the inverse is equal to g?
thanks for the help everyone!!
Applying what you've been given to the definition of a one-sided inverse, you can conclude the following:
1) From , g is acting as a left inverse of f and f is acting as a right inverse of g.
2) From , f is acting as a left inverse of g and g is acting as a right inverse of f.
Only a bijection, say f, has a unique two-sided inverse, i.e., a unique function that's both the left inverse and the right inverse of f.
But from 1) and 2), you have that g is both a left and right inverse of f, and that f is both a left and right inverse of g.
So on the one hand, you can say g = , the unique two-sided inverse of f.
On the other hand, you can say f = , the unique two-sided inverse of g.