Short exact sequence of vector spaces Suppose $V$ is a finite-dimensional vector space, and $U,W$ its subspaces. Is it possible to construct the following short exact sequence? $$0\rightarrow U\cap W\xrightarrow{f}U\oplus W\xrightarrow{g}U+W\rightarrow 0 \text{ (the direct sum here is external)}$$
I thought we could consider $U+W$ as $(U\oplus W)/(U\cap W)$ (the dimensions seem to permit it), then $f$ can be the standard injective inclusion $\iota$ of $U\cap W$ in $U\oplus W$ and $g$ the standard surjective "projection" $\pi$ onto the quotient space. Is this explicit enough a definition for $f$ and $g$? The exercise asks to construct these mappings, would there be any need to go into further detail than this?
 A: 
I thought we could consider U+W as (U⊕W)/(U∩W)

The main part of the exercise is to make this statement precise. So, yes, you need to go into further detail.
A dimension argument is not the intention, and in this case probably circular (in the sense that the proof of $\dim(U\cap W)+\dim(U+W)=\dim(U)+\dim(W)$ uses the result of this exercise).
It is also not a priori clear what the "standard inclusion of $U\cap W$ into $U\oplus W$" is.
Hint: Start by defining a surjective mapping $g:U\oplus V\rightarrow U+V$ and look at its kernel.
A: I think you can prove something stronger:
Let be $I$ and $J$ ideals of a ring $R$ ($R$-modules). Then the below sequence is exact:
$$
0 \to I \cap J \xrightarrow{f} I \oplus J \xrightarrow{g} I+J \to 0
$$
Indeed, you can identify $I \oplus J$ with $I \times J$. Then, you pick
$$
f: I \cap J \to I \oplus J \qquad x \mapsto (x,x)
$$
and
$$
g: I \oplus J \to I+J \qquad (x,y) \mapsto x-y
$$
The first function is invective obviously, while the second is surjective (nothing to say I think). Moreover, if $a \in Im (F)$, then there exists $x \in I \cap J$ such that $a = (x,x)$. Then $g(a) = g(x,x) = x-x=0$, then $a \in \ker (g)$. Viceversa, if $(x,y) \in \ker(g)$ then $x = y$. Thus $(x,y) \in Im(f)$.
Clearly, your exercise is a particular case of what I've just said (vector spaces over a field $K$ are just $K$-modules)
