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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Yes, range (f)=co-domain (f) implies surjection.
what ffm said
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)

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what sat said, in other words every element \(b \in B\) is the f-image of some element of A.
NOTE: If co-domain (f) \(\neq\) range (f) \( \implies\) Into function.
@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....
http://planetmath.org/encyclopedia/Similar.html
or look at "change of basis" here
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