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Math Help - Derivate that a function is injective

  1. #1
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    Derivate that a function is injective

    If we have that f and f o g (that is, f(g) ) are both injective, must then g be injective?

    I find that this is false and I show it with an example, where I have

    A = \{a,b,c,d\}, B=\{a,b,c,d\}, C=\{1,2,3,4\}
    g: A \Longrightarrow B, f: B \Longrightarrow C, f o g: A \Longrightarrow C

    g(a) = a, g(b) = b, g(c) = c, g(d) = c
    f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 4
    f(g(a)) = 1, f(g(b)) = 2, f(g(c)) = 3, f(g(d)) = 3

    Here I have that f and f o g are injective but g is not, hence g does not have to be injective.

    Is this a correct way to show this at all?
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  2. #2
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    Quote Originally Posted by posix_memalign View Post
    If we have that f and f o g (that is, f(g) ) are both injective, must then g be injective?

    I find that this is false and I show it with an example, where I have

    A = \{a,b,c,d\}, B=\{a,b,c,d\}, C=\{1,2,3,4\}
    g: A \Longrightarrow B, f: B \Longrightarrow C, f o g: A \Longrightarrow C

    g(a) = a, g(b) = b, g(c) = c, g(d) = c
    f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 4
    f(g(a)) = 1, f(g(b)) = 2, f(g(c)) = 3, f(g(d)) = 3

    Here I have that f and f o g are injective but g is not, hence g does not have to be injective.

    Is this a correct way to show this at all?


    No, it's not: as you showed,

    f\circ g(c)=f(g(c))=3=f(g(d))=f\circ g(d) \,,\,\,and\,\,c\neq d\Longrightarrow f\circ g is not 1-1.

    In fact, it is easy to show that

    if f\circ g is 1-1 then also g is 1-1 , and

    if f\circ g is onto then also f is.

    Try to prove the above.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    No, it's not: as you showed,

    f\circ g(c)=f(g(c))=3=f(g(d))=f\circ g(d) \,,\,\,and\,\,c\neq d\Longrightarrow f\circ g is not 1-1.

    In fact, it is easy to show that

    if f\circ g is 1-1 then also g is 1-1 , and

    if f\circ g is onto then also f is.

    Try to prove the above.

    Tonio
    I see now that I was wrong, thank you.

    I try again, I have that f and f o g are injective, it follows then that g is injective:

    I show this with a contradiction, assume that there exists elements s.t. g(a) = g(b) \Longrightarrow a \neq b and also assume that f and f o g are injective.
    Then we would have f(g(a)) = c = f(g(b)), where a \neq b
    This implies that f o g is not injective, which is a contradiction as it was assumed as a premise it was injective, hence g must be injective.

    Am I getting closer? :-)
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  4. #4
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    Quote Originally Posted by posix_memalign View Post
    I see now that I was wrong, thank you.

    I try again, I have that f and f o g are injective, it follows then that g is injective:

    I show this with a contradiction, assume that there exists elements s.t. g(a) = g(b) \Longrightarrow a \neq b and also assume that f and f o g are injective.
    Then we would have f(g(a)) = c = f(g(b)), where a \neq b
    This implies that f o g is not injective, which is a contradiction as it was assumed as a premise it was injective, hence g must be injective.

    Am I getting closer? :-)


    Yes, that was correct.

    Tonio
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  5. #5
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    A little correction.
    Quote Originally Posted by posix_memalign View Post
    I show this with a contradiction, assume that there exists elements s.t. g(a) = g(b) \Longrightarrow a \neq b
    It should say, "g(a) = g(b) and a\ne b".
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