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anonymous

  • 4 years ago

if f:A->B be a function and co-domain=range then the function is onto?

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  1. anonymous
    • 4 years ago
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    Yes, range (f)=co-domain (f) implies surjection.

  2. anonymous
    • 4 years ago
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    what ffm said

  3. anonymous
    • 4 years ago
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    range is set of all b in codomain such that there is an a in A with \[f(a)=b\] if that is equal to the codomain, then it means function is onto (surjective)

  4. anonymous
    • 4 years ago
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    what sat said, in other words every element \(b \in B\) is the f-image of some element of A.

  5. anonymous
    • 4 years ago
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    NOTE: If co-domain (f) \(\neq\) range (f) \( \implies\) Into function.

  6. anonymous
    • 4 years ago
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    @satellite and foolformath please help if u can. Similar matrices represent the same linear transformation. please prove it or send me a link of its prove its urgent....

  7. anonymous
    • 4 years ago
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    http://planetmath.org/encyclopedia/Similar.html

  8. anonymous
    • 4 years ago
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    or look at "change of basis" here

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