Thoughts on this Inverse Bijection Proof

[blindgibson27]

Is this sufficient?

 

[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.

 

[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.

How about this..

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 homework 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

 

[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.

 

[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.

 

[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 ZPrf
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