Proving Surjective and Injective Functions: f and g are Injections

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Homework Help Overview

The problem involves proving that two functions, f and g, are injections given that f is a surjection and the composition g∘f is an injection. The discussion centers around the properties of these functions in the context of set theory and function mappings.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the definitions of surjective and injective functions. There are attempts to clarify the reasoning behind proving the injectivity of f and g based on the properties of their composition.

Discussion Status

The discussion has progressed with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the assumptions made in the original attempts, and there appears to be a productive exchange of ideas about how to approach the proof for both functions.

Contextual Notes

Participants note the importance of the surjectivity of f and its role in the proof, as well as the need to avoid assuming properties of g without justification. There is an emphasis on ensuring that all definitions and properties are properly utilized in the reasoning process.

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Homework Statement



Let f: A→B be a surjection and g:B→C be such that g∘f is an injection. Prove that both f and g are injections

Homework Equations





The Attempt at a Solution



Suppose a = a' then g(f(a)) = g(f(a')) since g∘f is 1-1 we have that f(a) = f(a') hence f is 1-1.

Is this correct?
 
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You assumed that a=a'and proved that f(a) = f(a'), which is simply showing that f is a function, not anything to do with whether f is 1-1. Furthermore, you went from g(f(a)) = g(f(a')) to f(a) = f(a') which seems to assume that g is injective already.

You never used the fact that f is a surjection, which should also tip you off that you missed something.
 
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Let me redo.

Suppose a,a' ∈ A and that f(a) = f(a') then g(f(a)) = g(f(a')) since g∘f is an injection then we have a = a' therefore f is injective.

Is this correct?
 
That looks better. Any ideas on what to do for g?
 
Thanks! For g:

Let b, b' ∈ B. Then since f is onto there exists an a, a' ∈ A such that f(a) = b and f(a') = b'. Suppose g(b)=g(b') then we get g(b) = g(f(a)) = g(f(a'))= g(b') since g∘f is 1-1 then a = a' hence f(a) = f(a') therefore b = b' which means g is 1-1.

Is this correct?
 
That looks good to me.
 
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Thank you very much!
 

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