# Thread: Composition/Image Proof

1. ## 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 $f^-1(b)=a$
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.

2. Suppose $f(x) = y$.
Then $f^{-1}(y) = x$.

So $f^{-1}(f(x)) = f^{-1}(y) = x$.

Similarly,
$f^{-1}(y) = x$ gives us
$f(x) = y$.

So $f(f^{-1}(y)) = f(x) = y$.

3. Originally Posted by icemanfan
Suppose $f(x) = y$.
Then $f^{-1}(y) = x$.

So $f^{-1}(f(x)) = f^{-1}(y) = x$.

Similarly,
$f^{-1}(y) = x$ gives us
$f(x) = y$.

So $f(f^{-1}(y)) = f(x) = y$.
How does the im F=B work into the proof? I don't get that part.

4. 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, $f^{ - 1} (b) = \left\{ {x \in A:f(x) = b} \right\}$. That is the inverse image is a set.
The fact that $f$ is a bijection simply means that $\left( {\forall b \in B} \right)\left[ {f^{ - 1} (b) \ne \emptyset } \right]$.
By injectivity $\left\{ {p,q} \right\} \subseteq f^{ - 1} (b) \Rightarrow \quad f(p) = f(q) = b \Rightarrow \quad p = q$.
Which means that each $f^{-1}(b)$ is a singleton set.
So $\left( {\forall a \in A} \right)\left[ {f(a) \in B} \right] \Rightarrow \quad \left\{ a \right\} = f^{ - 1} \left( {f(a)} \right)$.

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

5. 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.

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