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proving composite function is injective if all it's constituent parts are injective Watch

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    I have so far

    I claim that the composition of injective functions is injective. Recall that injective means that \forall x,y \in \mathbb R , f(x)=f(y) \Rightarrow x=y.
    Let f(g(x))=f(g(y)) then using the definition of injective g(x)=g(y) and again using the definition of injective x=y therefore the composition of injective functions is injective.










    Ignore everything below
    But i'm not sure how to do like the general kind of proof.

    Like i know intuitively it is and i've tested a few but i'm not really sure how to prove it...

    I could do contraposition tho

    so

    P:=The composition of injective functions is injective
    Q:=An injective function
    ~Q:=Non-injective function


    i dunno


    direct proof?
    so if fog is injective then f and g are injective where f,g are in the set of ????

    I need some help
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    (Original post by will'o'wisp2)
    I have so far

    I claim that the composition of injective functions is injective. Recall that injective means that \forall x,y \in \mathbb R , f(x)=f(y) \Rightarrow x=y.
    Let f(g(x))=f(g(y)) then using the definition of injective g(x)=g(y) and again using the definition of injective x=y therefore the composition of injective functions is injective.










    Ignore everything below
    But i'm not sure how to do like the general kind of proof.

    Like i know intuitively it is and i've tested a few but i'm not really sure how to prove it...

    I could do contraposition tho

    so

    P:=The composition of injective functions is injective
    Q:=An injective function
    ~Q:=Non-injective function


    i dunno


    direct proof?
    so if fog is injective then f and g are injective where f,g are in the set of ????

    I need some help
    Yeah your proof is fine, I would just make the reasoning a bit more clear. You have g(x)=g(y) because f is injective and then you have x=y because g is injective.
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    (Original post by B_9710)
    Yeah your proof is fine, I would just make the reasoning a bit more clear. You have g(x)=g(y) because f is injective and then you have x=y because g is injective.
    ok thanks
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    (Original post by will'o'wisp2)
    ok thanks
    I assume you'll also want to prove that composition of surjective functions is surjective.
    Also remember that you can extend these results to any finite number of compositions of injective/surjective functions.
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    (Original post by B_9710)
    I assume you'll also want to prove that composition of surjective functions is surjective.
    Also remember that you can extend these results to any finite number of compositions of injective/surjective functions.
    woah there, hol up the master of prediction xD

    i can do an infinite regress with injective function but i'm not sure about surjective tho because the definition is slightly different but let me write up a rough answer of my injecive part and i'll take a look at the next one
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    (Original post by B_9710)
    I assume you'll also want to prove that composition of surjective functions is surjective.
    Also remember that you can extend these results to any finite number of compositions of injective/surjective functions.
    so then

    I claim that the composition of surjective functions is surjective. Recall that to be surjective means that  if f:X\Rightarrow Y then \forall y \in Y\ \exists x \in X : f(x)=y
    on second thought the stuff below don't work

    I've tried a few examples and it seem that let's say f(x)=x and g(x)=x+1 if we do f(g(x))=x+1 then g(f(x))=x+1 , it seems that the "bigger" of the 2 surjecive functions always seems to be the surjective one, so it seems to me in this case for example f(g(x))=g(x) and g(f(x))=g(x) which is surjective hence the composite of both is surjective but i'm not really sure where else to go with this.
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    (Original post by will'o'wisp2)
    so then

    I claim that the composition of surjective functions is surjective. Recall that to be surjective means that  if f:X\rightarrow Y then \forall y \in Y\ \exists x \in X : f(x)=y
    on second thought the stuff below don't work

    I've tried a few examples and it seem that let's say f(x)=x and g(x)=x+1 if we do f(g(x))=x+1 then g(f(x))=x+1 , it seems that the "bigger" of the 2 surjecive functions always seems to be the surjective one, so it seems to me in this case for example f(g(x))=g(x) and g(f(x))=g(x) which is surjective hence the composite of both is surjective but i'm not really sure where else to go with this.
    I'm not sure what you're saying here.
    If you have  f:A\rightarrow B and  g:B\rightarrow C that are both surjective, we want to show that  g\circ f :A\rightarrow C is surjective.
    So basically we want to show that for all z in C there exists an x in A such that  (g\circ f )(x)=z .
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    (Original post by B_9710)
    I'm not sure what you're saying here.
    If you have  f:A\rightarrow B and  g:B\rightarrow C that are both surjective, we want to show that  g\circ f :A\rightarrow C is surjective.
    So basically we want to show that for all z in C there exists an x in A such that  (g\circ f )(x)=z .
    i think i'll leave it there for today and come back tomorrow xD
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    (Original post by B_9710)
    I'm not sure what you're saying here.
    If you have  f:A\rightarrow B and  g:B\rightarrow C that are both surjective, we want to show that  g\circ f :A\rightarrow C is surjective.
    So basically we want to show that for all z in C there exists an x in A such that  (g\circ f )(x)=z .
    I have returned lol.

    So i handed my questions in and i think for my answer. I put.

    \forall c \in C\ \exists\ b \in B\ :\ f(b)=c \Rightarrow \exists\ a \in A\ :\ g(a)=b

    I think that works? That was the main argument anyway. I think i'll be ok, thanks a bunch for you help as always
 
 
 
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