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!
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),
So f is 1-1
Now to show it is onto,
so for each
there is a
so for each
(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!!
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.