Can you explain me why this is also isomorphism?

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    Explain Isomorphism
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Discussion Overview

The discussion revolves around the concept of isomorphism in the context of the logarithm function and its properties as a homomorphism. Participants explore the conditions under which the logarithm can be considered an isomorphism, particularly focusing on its behavior with positive numbers and the implications of bijectiveness.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant defines homomorphism and provides the logarithm function as an example, asking for clarification on its isomorphic nature.
  • Another participant states that an isomorphism is a bijective homomorphism, referencing external material for further explanation.
  • A participant questions the surjectiveness of the function ##f(x)=e^x## in relation to the isomorphism of the logarithm.
  • It is noted that the logarithm function can be considered an isomorphism if restricted to positive numbers.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which the logarithm function qualifies as an isomorphism, particularly regarding the necessity of restricting the domain to positive numbers. The discussion remains unresolved regarding the surjectiveness of the exponential function.

Contextual Notes

The discussion does not resolve the implications of restricting the logarithm to positive numbers or the conditions affecting the surjectiveness of the exponential function.

matematikuvol
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Homomorphism is defined by ##f(x*y)=f(x)\cdot f(y)##. One interesting example of this is logarithm function ##log(xy)=\log x+\log y##. Can you explain me why this is also isomorphism?
 
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I know that. But I asked only for logarithm because
[tex]\log (ab)=\log (a)+\log (b)[/tex]
[tex]\log ((-a)(-b))=\log (a)+\log (b)[/tex]
Why function ##f(x)=e^x## isn't surjective?
 


Ok. Tnx for the answer.
 

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