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    Note that id we know the image of a function and we consider it as f: Dom(f) \rightarrow Im(f) then it's automatically surjective


    what does the we consider it as f: Dom(f) \rightarrow Im(f) mean?
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    (Original post by will'o'wisp2)
    Note that id we know the image of a function and we consider it as f: Dom(f) \rightarrow Im(f) then it's automatically surjective


    what does the we consider it as f: Dom(f) \rightarrow Im(f) mean?
    Whenever you get stuck on a problem like this the first (really important) step is to make sure you understand all of the notation/definitions being used in the question. Break it into all of its separate parts and make sure you know every piece individually before worrying about the whole. Do you indeed know what Dom(f) and surjective are, for example?

    Anyway, when defining a function: f : A \rightarrow B

    We say f maps elements of the set A (known as the domain / dom(f)) onto elements of the set B (known as the codomain / cod(f)).

    Now, f is surjective if for every element b \in B there exists a \in A such that f(a) = b, i.e. every element of B can be "reached" from some element of A.

    Additionally, Im(f) is the subset of elements of B which can be reached from A.

    So, if we replace B with Im(f), and consider only the elements of B which can be reached from A, we suddenly notice (unsurprisingly) that every element of the new B (Im(f)) can now be reached from A. Therefore f : Dom(f) \rightarrow Im(f) is automatically surjective.
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    (Original post by President Snow)
    Whenever you get stuck on a problem like this the first (really important) step is to make sure you understand all of the notation/definitions being used in the question. Break it into all of its separate parts and make sure you know every piece individually before worrying about the whole. Do you indeed know what Dom(f) and surjective are, for example?

    Anyway, when defining a function: f : A \rightarrow B

    We say f maps elements of the set A (known as the domain / dom(f)) onto elements of the set B (known as the codomain / cod(f)).

    Now, f is surjective if for every element b \in B there exists a \in A such that f(a) = b, i.e. every element of B can be "reached" from some element of A.

    Additionally, Im(f) is the subset of elements of B which can be reached from A.

    So, if we replace B with Im(f), and consider only the elements of B which can be reached from A, we suddenly notice (unsurprisingly) that every element of the new B (Im(f)) can now be reached from A. Therefore f : Dom(f) \rightarrow Im(f) is automatically surjective.
    can you give an example? i don't quite understand because i know what surjective it and what makes a function so but i don't get what this is, is it supposed to be a special case or something?
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    (Original post by will'o'wisp2)
    can you give an example? i don't quite understand because i know what surjective it and what makes a function so but i don't get what this is, is it supposed to be a special case or something?
    Part of learning to be a mathematician is learning to make your own examples from pure theory. This isn't A-Level anymore and you cannot ask for an example every time something's a little difficult.

    Please read my post again - slowly and carefully - and as many times as is needed until the meaning begins to sink in. That is how mathematics works & the sooner you learn this incredibly valuable skill the better you'll do on your maths modules.

    Finally - please look up the definition of Im(f) and make sure you understand what that is too.
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    (Original post by President Snow)
    Part of learning to be a mathematician is learning to make your own examples from pure theory. This isn't A-Level anymore and you cannot ask for an example every time something's a little difficult.

    Please read my post again - slowly and carefully - and as many times as is needed until the meaning begins to sink in. That is how mathematics works & the sooner you learn this incredibly valuable skill the better you'll do on your maths modules.

    Finally - please look up the definition of Im(f) and make sure you understand what that is too.
    So i kind of understand now???

    So a function hasf:[0,2]=\mathbb R and f(x)=x^2 then this function is surjective because we know the domain maps onto the image?


    wait is it because the image is contained within the domain and so it's by definition surjective?

    so f:\mathbb R \rightarrow \mathbb Z is surjective because All integers are a part of the real numbers?
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    (Original post by will'o'wisp2)
    So i kind of understand now???

    So a function hasf:[0,2]=\mathbb R and f(x)=x^2 then this function is surjective because we know the domain maps onto the image?
    Not quite.

    The definition of a subjective map is as follows:

    The function f : X \rightarrow Y is surjective iff \forall y \in Y there exists x \in X such that f(x)=y


    Aside from me not understanding what you mean by "= R" part, I think you mean to say f : [0,2] \rightarrow \mathbb{R} with f(x)=x^2 - this is infact NOT surjective. The codomain is \mathbb{R}, so if I pick f(x)=9 then clearly there is no x in [0,2] such that f(x)=9.

    To fix your example, say f : [0,2] \rightarrow [0,4] - then this is surjective as for any f(x) \in [0,4] there is an x \in [0,2] which satisfies the map.

    If it's easier to understand, a map f: A \rightarrow B is surjective if f(A)=B
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    (Original post by RDKGames)
    Not quite.

    The definition of a subjective map is as follows:

    The function f : X \rightarrow Y is surjective iff \forall y \in Y there exists x \in X such that f(x)=y


    Aside from me not understanding what you mean by "= R" part, I think you mean to say f : [0,2] \rightarrow \mathbb{R} with f(x)=x^2 - this is infact NOT surjective. The codomain is \mathbb{R}, so if I pick f(x)=9 then clearly there is no x in [0,2] such that f(x)=9.

    To fix your example, say f : [0,2] \rightarrow [0,4] - then this is surjective as for any f(x) \in [0,4] there is an x \in [0,2] which satisfies the map.

    If it's easier to understand, a map f: A \rightarrow B is surjective if f(A)=B
    I think that's what i edited it to? You replied before you saw the edit, whoops


    so



    wait is it because the image is contained within the domain and so it's by definition surjective?

    so f:\mathbb R \rightarrow \mathbb Z is surjective because All integers are a part of the real numbers?
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    (Original post by will'o'wisp2)
    I think that's what i edited it to? You replied before you saw the edit, whoops


    so



    wait is it because the image is contained within the domain and so it's by definition surjective?

    so f:\mathbb R \rightarrow \mathbb Z is surjective because All integers are a part of the real numbers?

    Got derailed a bit when writing a response earlier...

    The codomain doesn't have to be contained within the domain for the function to be surjective - refer back to your own example that I fixed above.

    f : \mathbb{R} \rightarrow \mathbb{Z} is not necessarily surjective as it depends on what f is. Take f(x)= \lfloor x \rfloor which is a \mathbb{R} \rightarrow \mathbb{Z} mapping - this is surjective. Now take g(x)= \lfloor x \rfloor ^2 but this is not surjective since there is no x such that g(x)=-1 for example.
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    (Original post by RDKGames)
    Got derailed a bit when writing a response earlier...

    The codomain doesn't have to be contained within the domain for the function to be surjective - refer back to your own example that I fixed above.

    f : \mathbb{R} \rightarrow \mathbb{Z} is not necessarily surjective as it depends on what f is. Take f(x)= \lfloor x \rfloor which is a \mathbb{R} \rightarrow \mathbb{Z} mapping - this is surjective. Now take g(x)= \lfloor x \rfloor ^2 but this is not surjective since there is no x such that g(x)=-1 for example.
    so it's just a definiton of surjective then?
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    (Original post by will'o'wisp2)
    so it's just a definiton of surjective then?
    Referring to your OP, the "we consider it as " is in fact just working from the definition of surjectivity, yes. Clearly, if you know that the domain gets mapped onto the image, then everything in the image must have an element in the domain - this is the def.

    It's kinda like saying you start off with f : \mathbb{R} \rightarrow \mathbb{Z}, f(x)=\lfloor x \rfloor ^2 but then upon closer inspection you can narrow down the codomain such that it is f: \mathbb{R} \rightarrow \mathbb{N} \cup {0}. Now everything in the codomain \mathbb{N} \cup {0} (this is explicit image of f, since f can reach everything in this set, but not everything in \mathbb{Z}) has an element in \mathbb{R} hence fulfilling the definition. This means this mapping is now surjective.
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    (Original post by RDKGames)
    Referring to your OP, the "we consider it as " is in fact just working from the definition of surjectivity, yes. Clearly, if you know that the domain gets mapped onto the image, then everything in the image must have an element in the domain - this is the def.

    It's kinda like saying you start off with f : \mathbb{R} \rightarrow \mathbb{Z}, f(x)=\lfloor x \rfloor ^2 but then upon closer inspection you can narrow down the codomain such that it is f: \mathbb{R} \rightarrow \mathbb{N} \cup {0}. Now everything in the codomain \mathbb{N} \cup {0} (this is explicit image of f, since f can reach everything in this set, but not everything in \mathbb{Z}) has an element in \mathbb{R} hence fulfilling the definition. This means this mapping is now surjective.
    Oh right... i thought this was something super important in the lecture notes i made -__-

    Greatest thanks to you for explaining, and the other guy too.
 
 
 
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