
#1
Jan2013, 09:31 PM

P: 7

Is this sufficient? 



#2
Jan2013, 09:54 PM

P: 771

What are you trying to prove, exactly? These just look like definitions to me, in which case a much simpler description of a onetoone 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].




#3
Jan2113, 07:31 AM

P: 7

It is to proof that the inverse is a onetoone 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 



#4
Jan2113, 09:14 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Thoughts on this Inverse Bijection Proof
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]. 



#5
Jan2113, 12:01 PM

P: 7

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
Jan2113, 12:58 PM

P: 7

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] 


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