Thoughts on this Inverse Bijection Proof

blindgibson27

Is this sufficient?

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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 [itex]f:X \rightarrow Y[/itex] is an injection if, [itex]\forall a,b \in X, \ f(a) = f(b) \implies a = b[/itex].

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 [itex]f:X\rightarrow Y[/itex] be a one to one correspondence, show [itex]f^{-1}:Y\rightarrow X[/itex] is a one to one correspondence.

[itex] \exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2} [/itex]

furthermore, [itex]f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})[/itex] (by definition of function [itex]f[/itex] 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

HallsofIvy

Your proof that [itex]f^{-1}[/itex] 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 [itex]x\in X[/itex] then there exist [itex]y\in Y[/itex] such that [itex]f^{-1}(y)= x[/itex].

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

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 [itex]f:X \rightarrow Y[/itex] is onto [itex]Y[/itex], and [itex]g: Y \rightarrow Z[/itex] is onto [itex]Z[/itex], then [itex]g \circ f:X \rightarrow Z[/itex] is onto [itex]Z[/itex]


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