MHB Understand Theorem 2.15 - Bruce Cooperstein's Advanced Linear Algebra

[Math Amateur]

I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...

I am focused on Section 2.3 The Correspondence and Isomorphism Theorems ... ...

I need further help with understanding Theorem 2.15 ...

Theorem 2.15 and its proof read as follows:
https://www.physicsforums.com/attachments/5170In the above text ... in the proof of part (ii) we read the following:

" ... By i) above, it (the mapping being considered) is surjective ... ... "

My question is as follows:

How does it follow that the mapping being considered is surjective ... ?

Further, I am bothered that in part (i) we assumed T was surjective ... and this is not assumed in the proof of (ii) ... how then can we use part (i) as Cooperstein does? ... can someone please clarify this issue ...

Help with both the above issues/questions will be appreciated ...

Peter*** EDIT ***

A further concern to the above is the following:

Do we assume that the map T in part (ii) is a linear transformation ... I am assuming that we do ... is that right?*** EDIT 2 ***

Re-reading the Theorem I note that the introductory sentence is as follows:

"Let $$T \ : \ V \longrightarrow W$$ be a surjective linear transformation ... ... and this statement applies to the T in both (i) and (ii) ... which means that I have not been reading the Theorem carefully enough ... apologies if this (as I suspect) answers my problems above ... BUT ... if this is the case then surely in part (ii) Cooperstein should not have said " ... By i) above, it (the mapping being considered) is surjective ... ... " he should have said " ... By assumption, it (the mapping being considered) is surjective ... ... " ... ?

Peter

 

[Deveno]

There are "two" $T$'s.

The first is a linear transformation $T: V \to W$.

The second is a mapping from the SET of subspaces of $V$ that contain $\text{ker }T$ to the SET of subspaces of $W$.

With the first, we say $T(v) = w$ (the lower-case Latin letters are *elements*).

With the second, we say $T(V) = W$ (the upper-case Latin letters represent *vector spaces*).

Perhaps we should call the second mapping something like $\hat{T}$, and write, for a subspace $U$ of $V$:

$\hat{T}(U) = \{w \in W: w = T(u), u \in U\}$.

In any case, vector spaces are abelian groups, and the correspondence theorem for groups applies (since being abelian is a "global property" (that is, *all* elements commute with each other), any subgroup of an abelian group will be abelian, and any homomorphic image of an abelian group will *also* be abelian).

So all we have to do is verify that an abelian subgroup of $V$ that is closed under scalar multiplication gets mapped via a linear transformation to an abelian subgroup of $W$ that is also closed under scalar multiplication. This is easy to show:

Suppose $T: V \to W$ is linear, and that $U$ is an abelian subgroup of $V$ closed under scalar multiplication.

We need to show if $w \in T(U)$ and $a \in F$, that $aw \in T(U)$.

Since $w \in T(U), w = T(u)$ for some $u \in U$. Thus:

$aw = aT(u) = T(au)$. Since $au \in U$, then $aw \in T(U)$.

Since group homomorphisms (which every linear transformation is, and more) take identity to identity, we need not worry about $T(U)$ being empty, it will always contain the zero vector of $W$.

This is virtually the same correspondence theorem as any other, only the categories (and their associated morphisms) have changed.