Q:

Short Exact Sequence Vector Spaces

I've been posed the above question, but I'm a bit unclear what it's asking. Is $N$ any normal subgroup of $GL(U,V)$?  The homomorphism $\phi: GL(U,V) \to GL(U) \times GL(V/U)$ I plan on using is (I think?) the natural one $g \mapsto (g|U,\tilde{g})$ where $\tilde{g}: V/U \to V/U$ is the map $\tilde{g}(xU) = g(x)U$.
So my questions are (1): What is $N$? (2): Will this be the right homomorphism to use?
EDIT: also how to think about the isomorphism $N \simeq \text{Hom}(V/U,U)$? I understand intuitively why this is true I think, since if $g \in N$, $\phi(g) = (1_U,1_{V/U})$ which says $g(V/U) \subset U$ and $g(U) = U$. But what confuses me is how to construct an isomorphism where on one side we have linear map composition and on the other we have addition?

A:

In the finite dimensional case you can think in terms of matrices:
Choose a basis $u_1,...,u_d,u_{d+1}...,u_n$ of ( V ), such that the first $d$ form a basis of $U$.
In this basis, an element of $GL(U,V)$ is a block matrix
$$ \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} =: (A,B,C) $$
where the submatrix $A$ corresponds to the restriction to ( U ).
The multiplication law is $(A,B,C) \cdot (A',B',C') = (AA', AB'+BC',CC')$.
Your homomorphism $GL(U,V) \to GL(U)\times GL(V/U)$ is given by
$ (A,B,C) \mapsto (A,C) $ and the kernel consists of elements of the form $ (I,B,I) $, where $B$ is any $n-d\times d$ matrix. This corresponds to a map $Hom(V/U,U)$, because the $n-d$ vectors
$$ u_{d+1}+U,...,u_n+U $$
form a basis of $V/U$ and the $d$ vectors
$$ u_1,...,u_d$$
were a basis of $U$.
Surely a more abstract answer would be more satisfying, however, this view may be useful for you as well :)
PS: Allow me to use the same shortcut as above: The group law for elements of the form $(I,B,I)$ is $(I,B,I)\cdot (I,B',I) = (I,B+B',I)$, which explains the addition.

