Krull-Schmidt decomposition theorem states that for $R$-algebra $A$ and $M, N, M´$ and $ N´$ right $A$-modules such that $M = M_1 \oplus \ldots M_r $ and $N = N_1 \oplus \ldots N_s $, where $End_A(M_i)$ and $End_A(N_i)$ are local algebra (meaning that have unique maximal right ideal or have only 0 and 1 as its idempotents) for all $i,j$. If $M \simeq N, `$ then $r)s$ and there is a permutation $\sigma$ such that $M_i \simeq N_{\sigma(i)}.$
In the proof of this theorem by induction on $r$, in the final step we need to prove for $R$-algebra $A$ and $M, N, M´$ and $ N´$ right $A$-modules such that $M = M_1 \oplus M´$ and $N = N_1 \oplus N´$.
How can we prove that if $M´ \simeq N´$ by induction steps then $M_1 \oplus M´ \simeq N_1 \oplus N´$?
Or how to show $\frac{M_1 \oplus M´}{M_1} \simeq \frac{N_1 \oplus N´}{N_1} $ if we know $M´ \simeq N´$?
How can we use projections $\pi_1 : M_1 \oplus M´ \to M_1$ and $\bar{\pi}_1 : M_1 \oplus M´ \to N_1$ and $\pi´ : M_1 \oplus M´ \to M´$ and $\bar{\pi}´ : M_1 \oplus M´ \to N´$ and injections $\iota_1: M_1 \to M_1 \oplus M´ $ and $\bar{\iota}_1 : N_1 \to M_1 \oplus M´$ and $\iota´: M´ \to M_1 \oplus M´ $ and $\bar{\iota}´ : N´ \to M_1 \oplus M´$ for to do that?
Thanks!
When you say $(M_1 \oplus M')/M_1$, you have to be careful here, because $M_1$ is not literally a submodule of $M_1 \oplus M'$. Really, you are identifying $M_1$ as a submodule of $M_1 \oplus M'$ via $x \mapsto (x,0)$.
Projection onto the second component $M_1 \oplus M' \rightarrow M'$ is a surjective homomorphism with kernel $M_1$. The first isomorphism theorem induces an isomorphism $M' \cong (M_1 \oplus M')/M_1$. Similarly, $N' \cong (N_1 \oplus N')/N_1$. So assuming $M' \cong N'$, you get
$$(M_1 \oplus M')/M_1 \cong M' \cong N' \cong (N_1 \oplus N')/N_1$$