Composite Functions: Is (f o g)^-1 = gof?

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SUMMARY

The discussion centers on the composition of functions, specifically whether (f o g)^-1 equals gof. The consensus is that while f o g is a valid function, the statement (f o g)^-1 = gof is false. Instead, the correct relationship is (gof)^-1 = f^-1 o g^-1, indicating that the inverses of the composed functions must be taken in reverse order. This clarification highlights the importance of understanding function composition and inverses in mathematical contexts.

PREREQUISITES
  • Understanding of function composition (f o g)
  • Knowledge of inverse functions (f^-1 and g^-1)
  • Familiarity with domain and range concepts in mathematics
  • Basic principles of mappings between sets (A, B, C)
NEXT STEPS
  • Study the properties of function composition in detail
  • Learn about the conditions under which a function has an inverse
  • Explore examples of composite functions and their inverses
  • Investigate the implications of function domains and ranges in compositions
USEFUL FOR

Mathematics students, educators, and anyone interested in deepening their understanding of function theory and composition, particularly in advanced algebra or calculus contexts.

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If f:A -> B and g:B -> C are functions, is this true: f o g is also a function and (f o g) ^-1 = gof

I think this isn't true, but if this isn't the case, could someone please tell me a counter example?? Thanks
 
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the first part is true, but the send part is false. since it is an and statement the entire statement is false. if it were an or statement, it would be true.

I think I am right but I can't remember right now if the mapping from A to B gets changed when you take the inverse or what. either way, it should not be a function any more.
 
fog is not a function, composition reads from right to left. gof is a function.
 
Assuming f:A->B and g:B->C then gof is a function from A to C.

No, it is NOT true that (gof)-1= fog. As matt grime pointed out, that's not even a function: g goes from B to C and C is not the domain of f.

What IS true, in this case, is that (gof)-1= f-1og-1 which is probably what you meant.

Notice that, since f:A->B, f-1:B->A and, since g:B->C, g-1:C->B. That means that both (gof)-1 and f-1og-1 are from C to A.
 

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