Correspondence Theorem for Vector Spaces - Cooperstein Theorem 2.15

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SUMMARY

The discussion centers on Bruce Cooperstein's Theorem 2.15 from "Advanced Linear Algebra," specifically regarding the Correspondence and Isomorphism Theorems. Participants confirm that while the proof of part (i) does not depend on the linearity of the mapping, a linear transformation is necessary to ensure that the image of a vector space remains a vector space. The theorem asserts that for any surjective function \( f: A \to f(A) \), it holds that \( f(f^{-1}(Y)) = Y \) for any subset \( Y \subseteq f(A) \). This highlights the distinction between general functions and linear transformations in the context of vector spaces.

PREREQUISITES
  • Understanding of linear transformations and their properties
  • Familiarity with surjective functions and their implications
  • Knowledge of vector spaces and their structure
  • Basic grasp of the Correspondence and Isomorphism Theorems
NEXT STEPS
  • Study the proof of Theorem 2.15 in Bruce Cooperstein's "Advanced Linear Algebra"
  • Explore the implications of linearity in vector space mappings
  • Investigate the Correspondence Theorem in more depth
  • Learn about the properties of surjective functions in the context of set theory
USEFUL FOR

Students of linear algebra, mathematicians focusing on vector space theory, and educators teaching advanced algebra concepts will benefit from this discussion.

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I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...

I am focused on Section 2.3 The Correspondence and Isomorphism Theorems ... ...

I need help with understanding Theorem 2.15 ...

Theorem 2.15 and its proof read as follows:View attachment 5169It appears to me (and somewhat surprises me) that the proof of part (i) of the above theorem does not seem to depend on the linearity of T and hence would be true for any function/mapping f ...

But is my analysis correct ...

Could someone please confirm that I am correct ... or point out my error(s) ...

Peter
 
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For any surjective function:

$f: A \to f(A)$ it follows that for any subset $Y \subseteq f(A)$ that $f(f^{-1}(Y)) = Y$.

However, it takes a linear transformation to ensure the image of a vector space is again a vector space.
 
Deveno said:
For any surjective function:

$f: A \to f(A)$ it follows that for any subset $Y \subseteq f(A)$ that $f(f^{-1}(Y)) = Y$.

However, it takes a linear transformation to ensure the image of a vector space is again a vector space.

Oh! Excellent point ... I had not though of that ...

Thanks for the help ...

Peter
 

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