- #1
jgens
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Homework Statement
Suppose [itex]g[/itex] is a function with the property that [itex]g(x) \neq g(y)[/itex] if [itex]x \neq y[/itex]. Prove that there is a function [itex]f[/itex] such that [itex]f \circ g = I[/itex]
Homework Equations
A function is collection of ordered pairs with the property that if [itex](a,b)[/itex] and [itex](a,c)[/itex] are in the collection, then [itex]b = c[/itex].
The Attempt at a Solution
Here are my thoughts so far (this isn't a proof):
From the restrictions on [itex]g[/itex] we have that [itex]g(a) = g(b)[/itex] if and only if [itex]a = b[/itex]; hence, [itex]g[/itex] is the collection of ordered pairs [itex](x,g(x))[/itex] with the property that [itex](a,c)[/itex] and [itex](b,c)[/itex] are in the collection if and only if [itex]a = b[/itex].
If the composition of [itex]f[/itex] and [itex]g[/itex] is the identity function then [itex]f \circ g[/itex] is the collection of ordered pairs [itex]((x,g(x)),(f \circ g)(x)) = ((x,g(x)),x) = (g(x),x)[/itex] such that if [itex](a,b)[/itex] and [itex](a,c)[/itex] are in the collection, then [itex]b = c[/itex]. Since [itex]g[/itex] has this property, it seems like there should be some function [itex]f[/itex] such that [itex]f \circ g = I[/itex].
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!