quiLet $C$ be an abelian category and {$X_1$,...,$X_n$} a finite family of objects in that category. ( $X$,($M_i$: $X_i$$\to$ $X$) where $i_1$=1,....n a coproduct of the finite family if and only if there exist morphism $p_i$ :$X$$\to$$X_i$ ,i=1,...n such that
(a) $1_x$ = $M_i$$p_i$+....+$M_n$$p_n$ (b)$p_i$$M_i$ = $1_{X_i}$ , i=1,...n (c) $p_j$$M_i$= $O$ if i is different to j
My idea o prove the forward implication is to consider the diagram $0 \to X \to Y\to Z \to 0$ and $0 \to X \to X\oplus$Z$ \to Z \to 0$ where h: Y $\to$ X$\oplus$$Z$
I really need to end this proof . Thanks!!
You should just use the universal property, don't worry about exact sequences.
For starters assume the $p_i$ exist. You have to show that a family of maps $N_i\colon X_i \to Z$ factors through a unique map $X \to Z$. Well using the $p_i$ you get maps $X \overset{p_i}{\rightarrow} X_i \overset{N_i}{\rightarrow} Z$ so sum this over $i$ to get $X \to Z$. Then you'll have to use the properties of the $p_i$ to show that this is the only such map that all the $N_i$ factor through.
Secondly assume $X$ is a coproduct, you have to define $p_i$ satisfying those conditions. So fix $i$ and consider the family of maps $f_j\colon X_i \to X_j$ where $X_i \to X_i$ is the identity and $X_i \to X_j$ is zero when $i \neq j$. These maps, and the universal property of a coproduct, yield a map $p_i\colon X \to X_i$ through which the $f_i$ factor. Now you just have to prove that these $p_i$ satisfy the conditions that are listed.