Show that the function is 1-1 and onto

[evinda]

Hi! (Smile)

I want to prove that, if $f: A \xrightarrow[\text{onto}]{\text{1-1}} B$, then $f^{-1}: B \xrightarrow[\text{onto}]{\text{1-1}} A$.

That's what I have tried:

Let $x,y \in dom(f)$, with $x \neq y$.
Then, since $f:\text{ 1-1 }$, we have that $f(x) \neq f(y)$.
Also, since $f: \text{ onto } $, $\forall b \in B,\exists x$, such that $b=f(x)$.

Is it right so far?

We want to show that, if $y_1,y_2 \in B$, with $y_1 \neq y_2$, then $f^{-1}(y_1) \neq f^{-1}(y_2)$ and that $\forall x \in A, \exists d$, such that $f^{-1}(x)=d$, right?

If so, how could we show this?

 

[Fallen Angel]

Hi,

It's correct except the last equality, $$f^{-1}$$ should be applied in the rhs.

Given $$y \in B$$ and the condition f onto implies that $$f^{-1}(y) $$ is non empty, and 1-1 implies that $$f^{-1}(y)$$ is just one point so the inverse of f is well defined.

Try to reverse this argumentation.