I'm new to module theory, so this question may be really easy.
"Let $M$ be the direct sum of 2 $R$-submodule $M_1,M_2$ and $\pi_i (i=1,2)$ are the projections onto $M_i$. Let $N$ be a $R$-submodule of M. Prove $\pi_1(N)/(N\cap M_1) \cong \pi_2(N)/(N\cap M_2)$".
It looks symmetric, but I don't see the connection here. Beside the solution, can you show me the intuitive way to see it? Thank you.
Let $x\in N$, you can write $x=x_1+x_2, x_1\in M_1, x_2\in M_2$, define $f:\pi_1(N)/(N\cap M_1)\rightarrow \pi_2(N)/(N\cap M_2)$ by $f([x_1])=[x_2]$ where $[x_1]$ is the class of $x_1\in \pi_1(N)/(N\cap M_1)$.
$f$ is well-defined. If $x'\in N$, $x'=x'_1+x'_2, x_i\in M_i, i=1,2$, $[x'_1]=[x_1]$ implies that $x_1-x'_1\in N\cap M_1$, $x-x'=(x_1-x'_1)+(x_2-x'_2)$ implies that $x_2-x'_2=x-x'-(x_1-x'_1)\in N$ since $x-x'\in N$, we deduce that $x_2-x'_2\in N\cap M_2$ and $f([x'_1])=f([x_1])$.
$f$ is injective.
Suppose $f([x_1])=0$, this implies that $x_2\in N\cap M_2$, thus $x_1=x-x_2\in N, x_1\in N\cap M_1$ and $[x_1]=0$.
$f$ is surjective. For every $y\in \pi_2(N)/(N\cap M_2)$, $y=[x_2]$, $x=x_1+x_2, x\in N, x_i\in M_i$, $f([x_1])=[x_2]$.