In additive category product and co-product over finite family of objects are isomorphism.

Question

Let C be a additive category.Show that the co-product and product over finite family of objects are isomorphism.

Answer 1

Let $A_1,\ldots,A_n$ be objects, and let $P$ be the product, with projections $\pi_1,\ldots,\pi_n$. For every $j=1,\ldots,n$ define $i_j:A_j\to P$ to be the unique morphism that satisfies $\pi_j\circ i_j=id_{A_j}$ and $\pi_k\circ i_j=0$ for every $k\neq j$. ($i_j$ is well defined for every $j$ by the universal property of the product).

We show that $P$, together with the morphisms $i_1,\ldots,i_n$, satisfies the universal property of the coproduct. Let $B$ be another object, and for every $j$ let $f_j:A_j\to B$ be a morphism. Define $$f_P:P\to B$$ by $$f_P=\sum_jf_j\circ \pi_j.$$ Then for every $j$ $$f_P\circ i_j=\left(\sum_kf_k\circ\pi_k\right)\circ i_j=\sum_kf_k\circ \pi_k\circ i_j=f_j\circ\pi_j\circ i_j=f_j,$$ meaning that $f_j$ factors through $P$, and now we need to show the uniqueness of $f_P$. For that we first define $\varphi:P\to P$ by $$\varphi=\sum_ji_j\circ\pi_j,$$ and claim $$\varphi=id_P.$$To verify that it suffices, by the universal property of the product, to show $\pi_j\circ\varphi=\pi_j$ for all $j$. Indeed $$\pi_j\circ\varphi=\pi_j\circ\left(\sum_ki_k\circ\pi_k\right)=\sum_k\pi_j\circ i_k\circ\pi_k=\pi_j\circ i_j\circ \pi_j=\pi_j.$$Finally, assume that $f,g:P\to B$ satisfy $f\circ i_j=g\circ i_j$ for every $j$. Denote $h:=f-g$ and obtain $h\circ i_j=0$ for all $j$. But then $$h=h\circ id_P=h\circ\varphi=h\circ\sum_ji_j\circ\pi_j=\sum_jh\circ i_j\circ\pi_j=0,$$ and so $f=g$, and uniqueness is proven.