# Need help with second part of proof

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• Oct 10th 2012, 09:27 AM
franios
Need help with second part of proof
Suppose f:A→B is a given function. Then there is a function g:f(A)→A sich that
g o f=IA & f o g= I f(A), if and only if f is a one to one function and g=f-1
Proof(→):
Suppose f is 1-1 and g=f-1 then for all a elements of A, if f(a)=b then g(b)=a.
So g(f(a))=g(b)=a.
Thus g o f=IA, Similarly, f(g(b))=f(a)=b for any element b of f(A), so f o g=If(A).

Need help with (←)
• Oct 10th 2012, 10:03 AM
johnsomeone
Re: Need help with second part of proof
We're given $\displaystyle g \circ f = I_A$ and $\displaystyle f \circ g = I_{f(B)}$, and want to show that $\displaystyle f$ is 1-1, and then that $\displaystyle g = f^{-1}$.

What does a function being 1-1 mean? It means that if two images of the function are equal, then the originating domain elements must also be equal. (y = x^2 isn't 1-1, because a^2 = b^2 doesn't force a = b. y = x^3 is 1-1, since a^3 = b^3 does force a = b.)

Set it up, and then there's an obvious way to proceed. Like this:

Suppose $\displaystyle x_1, x_2 \in A$, and $\displaystyle f(x_1) = f(x_2)$.
Then... (fill in the reasoning)...
Therefore $\displaystyle x_1 = x_2$.
Therefore f is 1-1.

For $\displaystyle g = f^{-1}$, again, go to the definiton, which is that $\displaystyle f^{-1}(f(x)) = x \ \forall \ x \in A$.

Like this:

For all x in the domain, $\displaystyle g(f(x)) =$...(fill in the reasoning)... $\displaystyle =x$.
Therefore $\displaystyle g = f^{-1}$.
• Oct 10th 2012, 10:03 AM
FernandoRevilla
Re: Need help with second part of proof
Quote:

Originally Posted by franios
Need help with (←)

\displaystyle \displaystyle\begin{aligned}f(x_1)=f(x_2)& \Rightarrow g[f(x_1)]=g[f(x_2)]\\&\Rightarrow (g\circ f)(x_1)=(g\circ f)(x_2)\\&\Rightarrow I_A(x_1)=I_A(x_2)\\&\Rightarrow x_1=x_2\\&\Rightarrow f\mbox{ is injective}\end{aligned}

Try the rest.

P.S. Sorry, I didn't see the previous post.