I'll phrase your question in a different way - given is bijective, can you construct a function such that g is bijective and is the inverse of f?
Dear Friends ,
The following points can be proved easily :-
a) A function has a left inverse if and only if it is one-one.
b) A function has a right inverse if and only if it is onto.
A function is said to have an inverse if and only if its left and right inverses exist and are equal to each other.
We also know that a function has an inverse only if it is bijective. However the if part is never said. That is , it is never said that a function has an inverse if and only if it is bijective. We remain silent about the if part.
The question is that is it possible for a bijective function to have left and right inverses that are unequal ? Putting it into another way , the question is that is it possible for a function to be bijective and still not have an inverse ?
Please reply and point out any error in the argument if any .
Thanks
UJJWAL
@Defunkt ,
To be honest i did not understand your phrasing. I hope you would make it clearer. I would like to know how it can be phrased as such.
By the way I seem to have figured out the solution. I am putting it down.
Let me put down some basic results that can be proved easily :-
a) A function has got a left inverse iff it is one-one. Moreover, that left inverse is unique over the co-domain of the original function.
b) A function has got a right inverse iff it is surjective. Moreover that right inverse is unique over the co-domain of the original function.
Now suppose that we have a function f:X->Y. Suppose that f is bijective. So let g:Y->X be its left inverse and h:Y->X be its right inverse. Also suppsoe that there exists one yi in the set Y such that g(yi) is not equal to h(yi).
Now suppose g(yi)=xe1. It is easy to see that f(xe1)=yi otherwise g will not be the left inverse of f.
Also suppose that h(yi)=xe2.
Now foh (yi) = f(h(yi))=f(xe2) = yi (From the assumption that h is the right inverse of f ).
But f(xe1)= yi.
So we have f(xe1) = f(xe2) = yi and since f is bijective so we must have xe1 = xe2 . Hence our assumption is invalid and hence the left and the right inverse must be equal.
From here we conclude that if a function is bijective it will necessarily have a left inverse as well as a right inverse and they both shall necesarily be equal to each other, thus making the function possess a unique inverse.
I have just sketched this proof. Kindly go through it and feel free to pass any comments or suggestions that you may have.
Yours
UJJWAL
You askedI give you a bijective functionis it possible for a function to be bijective and still not have an inverse ?
That means that if I take any , there exists a unique such that .
That means that the inverse of f is well defined and exists for any , which is pretty much what you need.
I read through your proposed proof, the notation is a little unclear but I think it's alright.
Thank you for reading my proof. I understand it would be a bit unclear because i don't know about Latex so could not put it in a nice mathematical form. I am learning it and hopefully things would change.
I indeed needed the answer to whether i can construct the inverse function or not. The thing is that i needed the answer very rigorously. You know many people mistake the concept of an inverse function. The inverse of a function is defined in terms of the left inverse and the right inverse.
You know in the above post . I can define a number of f^{-1} all of which will be from B to A and will be well defined. But it is not necessary that all of them will be the inverse. I needed the answer that the left and the right inverse are necessarily equal when a function is bijective. I hope now I have got that. So my version and your version of the problem are different.
The definition I was using is an equivalent one - it does not have to be defined with left and right inverses.
Yes, but your function is which assigns a set to each element .You know in the above post . I can define a number of f^{-1} all of which will be from B to A and will be well defined. But it is not necessary that all of them will be the inverse. I needed the answer that the left and the right inverse are necessarily equal when a function is bijective. I hope now I have got that. So my version and your version of the problem are different.
The difference is that since f is injective, you can be assured that each will get matched with at most one .
And since f is surjective, you can be assured that each will get matched with at least one .
The conclusion follows..
It is almost impossible to read let alone follow your work.
Why not learn to post in symbols? You can use LaTeX tags.
[tex]f:A\to B[/tex] gives .
Frankly, I don’t understand your confusion with this.
Moreover, I don’t think you are justified in saying “it is never said that a function has an inverse if and only if it is bijective. We remain silent about the if part.” You do not know that is the case.
As a matter of definition: is a bijection if and only if
is injective and sujective.
Now if is a bijection then define .
Now it is completely obvious that each is a nonempty singleton set, in fact they partition the set .
That means that there is a perfectly natural way to define as .
The function is a bijection and is the right and left inverse of .