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    Basically the thread title.

    Does the fact that something is a mapping mean it is surjective by definition.

    If not can you give a counter example.

    If so, can you give an example of something that isn't surjective so I can see the point of the definition of surjection.

    Thanks
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    No, they're not. Take, for example,  f : \mathbb{R} \to \mathbb{R} , with  f(x) = x^2 . This map isn't surjective since, for example, -1 is not in the image of f.

    I think I see where your confusion is coming from. When a function is properly defined, a domain and a codomain are specified (as part of the definition). There's (often) a difference between the codomain of a function, and the image of a function. If  f : A \to B , then B is the codomain. The image would be written as  f(A) , which isn't necessarily the same as the codomain (as in my example above). The function is surjective when the image is the same as the codomain.

    The idea of codomain can seem a strange one. If  f : \mathbb{R} \to \mathbb{R} with  f(x) = x^2 , and  g : \mathbb{R} \to \mathbb{R}_0^+ with  g(x) = x^2 , then f and g are usually considered to be different functions. g is surjective, whereas f isn't.
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    I see what you mean. Thank you for the quick reponse and concise answer. Can I just clarify that  \mathbb{R}_0^+ is the non-negative real numbers.

    I have not seen that notation with the 0 and + before.

    Thanks.
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    (Original post by adie_raz)
    I see what you mean. Thank you for the quick reponse and concise answer. Can I just clarify that  \mathbb{R}_0^+ is the non-negative real numbers.

    I have not seen that notation with the 0 and + before.

    Thanks.
    No problem. Yes, that's what it means.
 
 
 
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