Proving Inverse Functions: g(x) \neq g(y) if x \neq y

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SUMMARY

The discussion centers on proving the existence of a function f such that f ∘ g = I, given that g is a function with the property that g(x) ≠ g(y) if x ≠ y. This property indicates that g is injective, meaning it maps distinct inputs to distinct outputs. The proposed solution involves defining f as the collection of ordered pairs that reverses the pairs in g, thereby satisfying the identity function requirement. The conclusion is that proving f is a function follows naturally from the definition of g.

PREREQUISITES
  • Understanding of function properties, specifically injective functions.
  • Familiarity with function composition and identity functions.
  • Knowledge of ordered pairs and their representation in mathematical functions.
  • Basic proof techniques in mathematics, particularly in function theory.
NEXT STEPS
  • Study the properties of injective functions in more detail.
  • Learn about function composition and its implications in mathematical proofs.
  • Explore the concept of identity functions and their role in function theory.
  • Investigate formal proof techniques for establishing function properties.
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Mathematics students, educators, and anyone interested in understanding function properties and proofs, particularly in the context of injective functions and function composition.

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

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