Composition of functions implies equality

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Homework Help Overview

The discussion revolves around the properties of bijective functions and their compositions. The original poster questions whether the equality of composed functions implies the equality of the functions themselves, given certain conditions on the functions involved.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to prove or disprove the statement regarding the equality of functions based on their compositions. Some participants explore the validity of counterexamples and the implications of relaxing conditions on the functions, such as considering surjectivity instead of bijectivity.

Discussion Status

The discussion is active, with participants presenting counterexamples and questioning the assumptions made about the functions. There is recognition of the need to clarify the conditions under which the original statement holds true.

Contextual Notes

Participants note that the original counterexample relies on specific properties of permutations and raises questions about the implications of changing the nature of the function h from bijective to surjective.

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We have three functions: f:A->A, g:A->A and h:A->A
with both f and g bijective and h bijective.

We know that f ° h = h ° g for every x in A.

Is it true that f=g for every x in A?

I have tried to solve it and I am pretty sure it is true but I cannot find neither a counterexample nor a simple proof.
 
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It took me just a few minutes to come up with a counter-example:

Let A= {1, 2, 3}, f= {(1,3), (2, 1), (3, 2)}, g= {(1, 2) , (2, 3), (3,1)}, h= {(1, 3), (2, 2), (3, 1)}.
 
Many thanks! Your solution is simple and hence great!
 
Your example is based on permutations..

What if h is only surjective rather than bijective?

The counterexample is not anymore valid...
 

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