Hello

I need some help to this problem...

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

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$\displaystyle b \in B$ s.t. g(b)=0 then q(g(b))=q(0)=0 while $\displaystyle q \circ g=h \circ v$ 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. $\displaystyle \text{Im} u= \text{Ker} v$ implying that $\displaystyle b \in \text{Ker}v=\text{Im} u.$ Therefore, there is an a $\displaystyle \in A s.t. u(a)=b.$

The left square is commutative, so $\displaystyle p \circ f= g \circ u.$ Then, p(f(a))=g(u(a))=g(b)=0. By the assumption f, p are injective, so is $\displaystyle f \circ p.$Hence, $\displaystyle p(f(a))=0 \rightarrow a=0$ 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 $\displaystyle b \in B$ so that v(b)=c. then [tex]h(v(b))=q(g(b)) => h(c)=q(g(b))=0 => $\displaystyle g(b) \in Ker q = im p$ => we have v $\displaystyle \in V s.t. p(v)=g(b).$

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!