Inverse Functions

1. Aug 12, 2009

jgens

1. The problem statement, all variables and given/known data

Suppose $g$ is a function with the property that $g(x) \neq g(y)$ if $x \neq y$. Prove that there is a function $f$ such that $f \circ g = I$

2. Relevant equations

A function is collection of ordered pairs with the property that if $(a,b)$ and $(a,c)$ are in the collection, then $b = c$.

3. The attempt at a solution

Here are my thoughts so far (this isn't a proof):

From the restrictions on $g$ we have that $g(a) = g(b)$ if and only if $a = b$; hence, $g$ is the collection of ordered pairs $(x,g(x))$ with the property that $(a,c)$ and $(b,c)$ are in the collection if and only if $a = b$.

If the composition of $f$ and $g$ is the identity function then $f \circ g$ is the collection of ordered pairs $((x,g(x)),(f \circ g)(x)) = ((x,g(x)),x) = (g(x),x)$ such that if $(a,b)$ and $(a,c)$ are in the collection, then $b = c$. Since $g$ has this property, it seems like there should be some function $f$ such that $f \circ g = I$.

Now, I'm sure that what I've shown so far is an abuse of/improper notation and probably is not valid. I would just like some help to string everything together. Thanks!

2. Aug 12, 2009

slider142

Since you are defining a function as a collection of ordered pairs, why not define f to be the collection of ordered pairs that corresponds to reversing the order of the pairs in the collection g(x). Then the fact that fog = I is by definition, and you have only to prove that f is a function, which should be automatic given the definition of g.

3. Aug 12, 2009

jgens

I actually thought about that about a half-hour after my last post. Thanks!