Let $R$ be a ring, $M$ and $R$-left module and $U, V$ submodules of $M$. I want to show that
$$f: M/U \cap V \to M/U \oplus M/V, m + U \cap V \mapsto (m + U, m + V)$$
and
$$g: M/U \oplus M/V \to M/(U+V), (m + U, n + V) \mapsto m - n + U + V$$
are two well-defined linear transformations, and that f is injective, g is surjective and $\ker g = Im f$.
Thanks in advance. I'm not quite used to working with constructions like these, so any help would be appreciated.
On
Well, the first is clearly well-defined, for if $m_1 + (U\cap V)$ and $m_2 + (U \cap V)$ are representatives of the same class, then $m_1 - m_2 \in (U \cap V) \subset U$, so $m_1 + U$ and $m_2 + U$ represent the same class in $M/U$, and a similar argument works for $V$.
Can you run with that to show that the second is well-defined?
I'll leave well-defined and linear to you since I don't like to show those two. To show $f$ is injective, note if $x\in M/(U\cap V)$ with $f(x)=(0,0)$ then $x\in U$ and $x\in V$. So $x\in U\cap V$, i.e. $x=0\in M/(U\cap V)$.
To show $g$ is surjective, note for $x\in M/(U+V)$ we can take $(x+U,0+V)$ so that $f(x+U,0+V)=x+U+V$.
Take an element $m+(U\cap V)$. Then $f(m+(U\cap V))=(m+U, m+V)$. Now $g(m+U, m+V)=m-m+U+V= U+V$ so that $\operatorname{im}(f)\subset\ker(g)$. And you need to show the reverse inclusion.