I need some help to this problem...
We say that a sequence of morphisms of K-vector spaces is an exact sequence if
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 s.t. g(b)=0 then q(g(b))=q(0)=0 while 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. implying that Therefore, there is an a
The left square is commutative, so Then, p(f(a))=g(u(a))=g(b)=0. By the assumption f, p are injective, so is Hence, 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 so that v(b)=c. then [tex]h(v(b))=q(g(b)) => h(c)=q(g(b))=0 => => we have v
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...