vector space

Printable View

January 7th 2012, 01:04 PM
ely_en

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)

http://img819.imageshack.us/img819/4417/92539387.jpg

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!(Happy)
January 8th 2012, 06:36 AM
Opalg

Re: vector space

Quote:

Originally Posted by ely_en

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).
January 8th 2012, 10:23 AM
ely_en

Re: vector space

thank you! the problem is solved now!