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  1. #1
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    vector space

    Hello
    I need some help to this problem...


    We say that a sequence of morphisms of K-vector spaces V_1--f-->V_2--g-->V_3 is an exact sequence if Ker(g) = Im(f).

    We have this diagram of K-vector spaces in which the horizontal lines are exact sequences and we also know that the squares are commutative. ( pof=gou , qog=hov)



    a) Prove that if f, h, p are injections, then g is a injection.
    b) State and prove a similar result with the one from a), on the same diagram.

    Thank you so much!

    This is the proof for a) : Let b \in B s.t. g(b)=0 then q(g(b))=q(0)=0 while q \circ g=h \circ v so q(g(b))=h(v(b))=0. By hypothesis, h is injective, so v(b)=0, also the first row is exact i.e. \text{Im} u= \text{Ker} v implying that b \in \text{Ker}v=\text{Im} u. Therefore, there is an a \in A s.t. u(a)=b.

    The left square is commutative, so p \circ f= g \circ u. Then, p(f(a))=g(u(a))=g(b)=0. By the assumption f, p are injective, so is f \circ p.Hence, p(f(a))=0 \rightarrow a=0 implying that b=u(a)=u(0)=0. Therefore, the kernel of g is trivial, so g is injective.

    But for b), how can I prove for example that "if p is injective and v is surjective, f is surjective and g is injective => h is injective, too."

    I tried this way..
    Let c so that h(c)=0.
    v is surjective, so we have b \in B so that v(b)=c. then [tex]h(v(b))=q(g(b)) => h(c)=q(g(b))=0 => g(b) \in Ker q = im p => we have v \in V  s.t. p(v)=g(b).
    We know that Ker q= Imp => q(p(v))=0
    But I don't know how to continue from this point..

    it must be something like in 5 lemma's proof...

    thank you!
    Last edited by ely_en; January 7th 2012 at 03:02 PM.
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  2. #2
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    Re: vector space

    Quote Originally Posted by ely_en View Post
    Hello
    I need some help to this problem...


    We say that a sequence of morphisms of K-vector spaces V_1--f-->V_2--g-->V_3 is an exact sequence if Ker(g) = Im(f).

    We have this diagram of K-vector spaces in which the horizontal lines are exact sequences and we also know that the squares are commutative. ( pof=gou , qog=hov)

    \begin{array}{ccccc} A\ & \overset{u}{\longrightarrow} & B\ & \overset{v}{\longrightarrow} & C\ \\ \downarrow\,\scriptstyle f && \downarrow\,\scriptstyle g && \downarrow\,\scriptstyle h \\ V\ & \overset{p}{\longrightarrow} & W\ & \overset{q}{\longrightarrow} & T\ \end{array}

    a) Prove that if f, h, p are injections, then g is an injection.
    b) State and prove a similar result with the one from a), on the same diagram.
    My suggestion for b) would be to take the statement of a) and "turn it upside down", as follows:

    b) Prove that if f, h, v are surjections, then g is a surjection.

    You can prove this by a similar diagram-chasing procedure to that used for a).
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  3. #3
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    Re: vector space

    thank you! the problem is solved now!
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