My thoughts are that f^{-1} cancels by definition,
so to begin let f^{-1}(x_{1}) = f^{-1}(x_{2})
\Leftrightarrow f(f^{-1}(x_{1})) = f(f^{-1}(x_{2})) since \forall y \in Y the function f^{-1}(y) = x for some x \in X
\Leftrightarrow x_{1} = x_{2}
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 =...
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.
Homework Statement
Prove that the inverse of a bijective function is also bijective.
Homework Equations
One to One
f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2}
Onto
\forall y \in Y \exists x \in X \mid f:X \Rightarrow Y
y = f(x)
The Attempt at a Solution
It...
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...