$\exists N\leq M$ such that $K_{i}=N\cap M_{i}$ and $N_{i}=\pi_{i}\left(N\right)$($i=1,2$) ($\pi$ is projection) Let $M$ be direct sum of two submodules $M_{1},M_{2}$. Prove that if $K_{i}\leq N_{i}\leq M_{i}$ ($...\leq...$ : $...$ be submodule of $...$) and $N_{1}/K_{1}\cong N_{2}/K_{2}$ then $\exists N\leq M$ such that $K_{i}=N\cap M_{i}$ and $N_{i}=\pi_{i}\left(N\right)$($i=1,2$)
($\pi$ is projection)
 A: It took some guessing,  but you know you can $\cap$ and $\oplus$, when the intersection is $\{0\}$, to make new spaces, and we can also $+$ but that wasn't used.  I got lucky and found this solution.
Let $N = (N_1 \oplus K_2) \cap (N_2 \oplus K_1)$.
I assume $\pi_i : M_1 \oplus M_2 \to M_i$ is the projection homomorphism sending $m_1 + m_2$ to $m_i$.  We need to check that $\pi_i (N) = N_i$.  Now, notice that each $n\in N$ can be written $n = n_1 + k_2 = n_2 + k_1$, because at the intersection that's what happens.  So we can choose which ever formula is most convenient.  So $\pi_i(n) = n_i$, where $n = n_i + k_j$.   We need projection onto $N_i$, so choose $n_i \in N$.  Notice that $n_i + k_j = n_j + k_i$, so that $n_i = (n_j + k_j) + k_i$.  Since $K_i \leqslant N_i$, we have that $n_i \in N_j \oplus K_i$. We alread know that $\pi_i(N_i) \subset N_i$ by definition of the projection.  So we have that
$$
\forall n_i \in N_i, \ \exists n_j \in N_j, k_i\in K_i \text{ such that } n_i = n_j + k_i. 
$$
so then $\pi_i (n_i + k_j) = \pi_i(n_i) = \pi_i(n_j + k_i) = n_i$, so clearly $n_i$ is in $\pi_i(N)$, since $\pi_i(N) = \pi_i(N_1 \oplus K_2) \cap \pi_i(N_2 \oplus K_1)$.  Thus we've shown the reverse set inclusion and we're done.
Consider $(N_1 \oplus K_2) \cap M_1$ .  If $n_1 + k_2 \in N_1 \oplus K_2$ and $n_1 + k_2 = m_1 \in M_1$.  Then $k_2 = m_1 - n_1 \in M_1$, so $k_2$ = 0.  Thus the intersection equals $N_1$.  Similarly $(N_2 \oplus K_1) \cap M_1 = K_1$.  And so the whole intersection $N \cap M_i = K_i$ for each $i$.
QED
