Thread: Composition/Image Proof

If f:A→B is 1-1 and im f=B prove that (f^-1 ○f)(a)=a for all a element of A and (f^-1 ○f)(b)=b for each b an element of B.


Here is what I've got out of this.
f is 1-1, so for each b, there is one and only a. So
Im f=b. Thus, f is onto. Therefore, the range is equal to the set B. Would this imply f(a)=b and f(b)=b?

I'm just not sure where to go with this problem.

Suppose .
Then .

So .

Similarly,
gives us
.

So .

Originally Posted by icemanfan

Suppose .
Then .

So .

Similarly,
gives us
.

So .

How does the im F=B work into the proof? I don't get that part.

Originally Posted by kathrynmath

How does the im F=B work into the proof? I don't get that part.

One reason I did not respond before now is that I consider this question to have notational problems. Those problems are reflected in the icemanfan’s answer, he assumed the ordinary meaning of the symbols.

However is these set theory proofs, . That is the inverse image is a set.
The fact that is a bijection simply means that .
By injectivity .
Which means that each is a singleton set.
So .

Now I understand that this notation is not what the question implies.
However, I hope it may give you some guidance.

Originally Posted by kathrynmath

How does the im F=B work into the proof? I don't get that part.

That just means that all of the elements of B are values that can be output by F. Hence, F is a bijection, and all elements of B have inverses of F in A.

Thanks, I think I got the right idea to finish the problem now.