# Thoughts on this Inverse Bijection Proof

1. Jan 20, 2013

### blindgibson27

Is this sufficient?

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2. Jan 20, 2013

### Number Nine

What are you trying to prove, exactly? These just look like definitions to me, in which case a much simpler description of a one-to-one function would be as follows: A function $f:X \rightarrow Y$ is an injection if, $\forall a,b \in X, \ f(a) = f(b) \implies a = b$.

3. Jan 21, 2013

### blindgibson27

It is to proof that the inverse is a one-to-one correspondence. I think I get what you are saying though about it looking as a definition rather than a proof.

Let $f:X\rightarrow Y$ be a one to one correspondence, show $f^{-1}:Y\rightarrow X$ is a one to one correspondence.

$\exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2}$

furthermore, $f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})$ (by definition of function $f$ and one to one)

kind of stumped from this point on..
I may want to transfer this post over to the hw section though, I did post to just get a confirmation on my thoughts on bijection but it is now turning into something a bit more specific than that

4. Jan 21, 2013

### HallsofIvy

Your proof that $f^{-1}$ is injective is correct. Your proof that it is surjective does not look to me like it actually says anything!

You want to prove that, if $x\in X$ then there exist $y\in Y$ such that $f^{-1}(y)= x$.

Given $x\in X$, let $y= f(x)$. Then it follows that $f^{-1}(y)= f^{-1}(f(x))= x$.

5. Jan 21, 2013

### blindgibson27

That makes a lot of sense and I am following that thought process, thank you for clearing that up for me. I was just stumped on the direction of the onto.

6. Jan 21, 2013

### blindgibson27

In the same light, these are my thoughts on my next exercise. If I have this wrong I may need to solidify my idea on the concept a bit more.

It reads: Show that if $f:X \rightarrow Y$ is onto $Y$, and $g: Y \rightarrow Z$ is onto $Z$, then $g \circ f:X \rightarrow Z$ is onto $Z$

Prf
Given $y \in Y$, let $y = g^{-1}(z)$ and $x = f^{-1}(y)$
$\forall z \in Z$, $f^{-1}(g^{-1}(z)) = f^{-1}(y) = x$