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=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 (←)

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 $\displaystyle g = f^{-1}$.

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.