Need help with second part of proof

**franios** · October 10th 2012, 09:27 AM

Suppose f:A→B is a given function. Then there is a function g:f(A)→A sich that
g o f=I_A & 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=I_A,Similarly, f(g(b))=f(a)=b for any element b of f(A), so f o g=I_f(A).

Need help with (←)

**johnsomeone** · October 10th 2012, 10:03 AM

We're given $g \circ f = I_A$ and $f \circ g = I_{f(B)}$ , and want to show that $f$ is 1-1, and then that $g = f^{-1}$ .

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

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

Like this:

For all x in the domain, $g(f(x)) =$ ...(fill in the reasoning)... $=x$ .
Therefore $g = f^{-1}$ .

**FernandoRevilla** · October 10th 2012, 10:03 AM

Originally Posted by franios

Need help with (←)

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

Try the rest.

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

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