$F \times G$ as defined is indeed a product in $A^I$, $ \ A $ an additive category.

Question

Trying to prove: $F \times G$ is a product in $A^I$. It has been defined as $(F\times G)(i) = F(i) \times G(i)$ and $(F\times G)(f) = $ unique map $F(i) \times G(i) \to F(j) \times G(j)$ described in the UMP of products in $A$.

$\pi^F : F \times G \to F$, sim $G$, are natural maps and for each $i \xrightarrow{f} j$ in $I$, $\pi^F_i : F(i) \times G(i) \to F(i)$, sim $G$, are such that for any natural maps $\eta^F: H \to F$, $ \ \eta^G : H \to G$ there exists a family of maps in $A$, $\eta : H \to F \times G$ such that $\pi^F \eta = \eta^F$, $ \ \pi^G \eta = \eta^G$.

How do I show that $\eta$ is a natural map?

Answer 1

To prove the naturality of $\eta$ it is enough to exhibit for an arbitrary arrow $f:i\to j$ in $I$ that: $$\left(Ff\times Gf\right)\circ\eta_{i}=\eta_{j}\circ Hf$$

We have:

$$\pi_{j}^{F}\circ\left(Ff\times Gf\right)\circ\eta_{i}=Ff\circ\pi_{i}^{F}\circ\eta_{i}=Ff\circ\eta_{i}^{F}=\eta_{j}^{F}\circ Hf=\pi_{j}^{F}\circ\eta_{j}\circ Hf$$

The first equality as a consequence of the naturality of $\pi^{F}$ and the third as a consequence of the naturality of $\eta^{F}$.

Likewise we find:

$$\pi_{j}^{G}\circ\left(Ff\times Gf\right)\circ\eta_{i}=Ff\circ\pi_{i}^{G}\circ\eta_{i}=Ff\circ\eta_{i}^{G}=\eta_{j}^{G}\circ Hf=\pi_{j}^{G}\circ\eta_{j}\circ Hf$$

on base of the naturality of $\pi^{G}$ and $\eta^{G}$.

Source $\left(\pi_{j}^{F},\pi_{j}^{G}\right)$is a mono-source so we are allowed to conclude that indeed: $$\left(Ff\times Gf\right)\circ\eta_{i}=\eta_{j}\circ Hf$$

Answer 2

Thanks @drhab for your explanation. I will try to elaborate here for understanding.

$\pi_i^F : F(i) \times G(i) \to F(i)$ is monic

If we have: $$ k \xrightarrow{e_1, e_2} F(i) \times G(i) \xrightarrow{\pi^F_i} F(i) \\ k \xrightarrow{e_1, e_2} F(i) \times G(i) \xrightarrow{\pi^G_i} G(i) $$

Then by definition of product in $A$, there exists a unique map $k \xrightarrow{e} F(i) \times G(i)$ such that $\pi_i^F e = (\pi_i^F e_1 = \pi_i^F e_2)$, sim $G$, $\implies e_2 = e = e_1$, sim $G$.