Basic properties of additive categories

Question

I am trying to understand additive categories but I have trouble with basic properties.

First, want to show that in an additive category $A$ the product $M \times N$ is isomorphic to the direct sum $M \oplus N$. For construct a map $M \times N \to M \oplus N$, one can take $i_M \circ \pi_M + i_N \circ \pi_N$. For the other direction we can take the sum $(id_M,0)\oplus(0,id_N)$ gives us a map : $M \oplus N \to M \times N$. I don't see why these maps should be inverse.

Secondly, I saw that a functor $F : A \to A'$ between two additive categories is "$\mathbb Z$ linear on the morphisms" (i.e $F(f+g) = F(f) + F(g)$) if and only if the natural map $F(M) \oplus F(L) \to F(M \oplus L)$ is an isomorphism. I also don't see why.

Many thanks in advance !

Answer 1

$\renewcommand{\id}{\text{id}} $Given $\varphi \colon M \oplus N \to M \times N$ and $\psi \colon M \times N \to M \oplus N$, you get a map $\psi\circ\varphi \colon M \oplus N \to M \oplus N$. You can show that $$\psi \circ \phi \circ i_M = i_M = \id_{M \oplus N} \circ i_M \colon M \to M \oplus N$$ and $$\psi \circ \phi \circ i_N = i_N = \id_{M \oplus N} \circ i_N \colon N \to M \oplus N.$$ By the uniqueness property of the coproduct, this implies that $\psi \circ \varphi = \id_{M \oplus N}$. You can do a similar thing to show that $\varphi \circ \psi = \id_{M\times N}$.

For the second part, $F$ gives you two natural maps $F(M) \oplus F(L) \to F(M \oplus L)$ and $F(M \times L) \to F(M) \times F(L)$. By the argument above, you can use that $F$ is $\mathbb{Z}$-linear to show that these are inverses (when viewed through the isomorphism between product and coproduct). For the other direction, if $F \colon A \to A'$ preserves coproducts, then given $f,g \colon M \to N$ in $A$ you can consider the maps $$ F(M) \xrightarrow{F\begin{pmatrix}1\\1\end{pmatrix}} F(M \oplus M) \xrightarrow{F(f \; g)} F(N)$$ and $$ F(M) \xrightarrow{\begin{pmatrix}1\\1\end{pmatrix}} F(M) \oplus F(M) \xrightarrow{(F(f) \; F(g))} F(N)$$ and show that the first is $F(f+g)$, the second is $F(f)+F(g)$, and that they are equal.

(Note: I'm using a kind of matrix notation which is pretty handy; given maps $a_{ji} \colon M_i \to N_j$, I can write $(\begin{smallmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{smallmatrix}) \colon M_1 \oplus M_2 \to N_1 \oplus N_2$, and composition of such morphisms corresponds to the usual matrix multiplication.)

Answer 2

Hint: Use the universal property of the sum.

Let $0$ be the nul element. You have morphisms $Id_M:M\rightarrow M, u_M:M\rightarrow 0\rightarrow N$ where $0$ is the zero object. The universal property of the product implies the existence defines $i_M:M\rightarrow M\times N$ such that $p_M\circ i_M=Id_M, p_N\circ i_M=u_M$.

Now consider morphisms $f:M\rightarrow W, g:N\rightarrow M$ consider $h=f\circ p_M+g\circ p_N, h\circ i_M=f, h\circ i_N=g$. If $h':M\times N\rightarrow W$ such that $h'\circ i_M$ factors by $0_W$ and, $h'\circ i_N$ factors by $0_W$ where $0_M:0\rightarrow M$, then by the universal property of the product, $h'=0$. This implies that $h$ is unique, so $M\times N$ is the sum.