exsistance of inverse?
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!
MacstersUndead, when talking about functions, the "inverse" always refers to the inverse function, the function g(x) such that f(g(x))= x and g(f(x))= x, NOT the "reciprocal". A constant function does NOT have an inverse.
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!!
What properties does an inverse function have?
What properties does g have?
Show that the inverse of f is the unique function such that and . Since you are told that g has those properities, it [b]is[/b the inverse.
Applying what you've been given to the definition of a one-sided inverse, you can conclude the following:
Originally Posted by matt.qmar
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.