# bijection

• Sep 6th 2008, 10:48 AM
dori1123
bijection
It is given that
(1) f: A --> B is injective iff f has a left inverse. (f has a left inverse if there is a function g: B --> A such that g(f(a)) = a for all a in A)
(2) f is surjective iff f has a right inverse. (f has a right inverse if there is a function g: B --> A such that f(g(b)) = b for all b in B)

I need to prove that
f is a bijection iff there exists g: B --> A such that f(g(b)) = b for every b in B and g(f(a)) = a for every a in A.

So if I am given (1) and (2) above, can I prove it like this?
i. Suppose f is bijective. That is, f is injective and surjective. Then by (1) there is a function g: B --> A such that g(f(a)) = a for every a in A since f is injective, and by (2) f(g(b)) = b for every b in B. Thus, g is the inverse of f.
ii. Suppose f has an inverse g: B --> A such that f(g(b)) = b for every b in B and g(f(a)) = a for every a in A. Then by (1) and (2) f is injective and surjective. Therefore, f is bijective.

Will this work?
• Sep 6th 2008, 05:31 PM
ThePerfectHacker
Quote:

Originally Posted by dori1123
It is given that
(1) f: A --> B is injective iff f has a left inverse. (f has a left inverse if there is a function g: B --> A such that g(f(a)) = a for all a in A)

Let me do this one and you try the other one.

Say $f: A\to B$ is injective. Define a function $g:B\to A$ in the following way: let $b\in B$ if $b=f(a)$ for $a\in A$ then define $g(b) = a$ otherwise define $g(b) = b$. It is important to show this is well-defined meaning it makes sense how we defined this function. Perhaps there are several values of $a$ such that $b=f(a)$ - if so then which one do we choose for $g(b) = a$ ? This is not a problem because the function is injective thus these values of $a$ are unique. This defines a function $g: B\to A$. Now we need to confirm that $g(f(a)) =a$. And this is of course true on how we defined $g$ above.

You try the converse and the second part.