# Math Help - Function and its inverse

1. ## Function and its inverse

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

2. I'll phrase your question in a different way - given $f: A \to B$ is bijective, can you construct a function $g:B \to A$ such that g is bijective and is the inverse of f?

3. @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

is it possible for a function to be bijective and still not have an inverse ?
I give you a bijective function $f: A \to B$

That means that if I take any $b \in B$, there exists a unique $a \in A$ such that $f^{-1}(b) = a$.
That means that the inverse $f^{-1}$ of f is well defined and exists for any $b \in B$, 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.

5. 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.

6. Originally Posted by UJJWAL
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.
The definition I was using is an equivalent one - it does not have to be defined with left and right inverses.

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.
Yes, but your function is $f^{-1}: B \to \mathbb{P}(A)$ which assigns a set to each element $b \in B$.
The difference is that since f is injective, you can be assured that each $b \in B$ will get matched with at most one $a \in A$.
And since f is surjective, you can be assured that each $b \in B$ will get matched with at least one $a \in A$.
The conclusion follows..

Why not learn to post in symbols? You can use LaTeX tags.
$$f:A\to B$$ gives $f:A\to B$.

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: $f:A\to B$ is a bijection if and only if
$f$ is injective and sujective.

Now if $f:A\to B$ is a bijection then $\left( {\forall t \in B} \right)$ define $A_t = f^{ - 1} \left( {\{ t\} } \right) = \left\{ {x \in A:f(x) = t} \right\}$.
Now it is completely obvious that each $A_t$ is a nonempty singleton set, in fact they partition the set $A$.
That means that there is a perfectly natural way to define $g:B\to A$ as $t\mapsto x\in A_t$.
The function $g$ is a bijection and is the right and left inverse of $f$.