I work in a category $\mathbb{C}$, and use definitions and notation from the book 'Category Theory', by Steve Awodey.
I'm learning some basic category theory from Awodey's book as part of self studying commutative algebra. So, I'm very new at this. In trying to understand naturality, I tried to see the simple and "natural" isomorphism between $A\times B$ and $B\times A$. I reason as follows:
Given the two products, $A\times B$ and $B\times A$, they already come equipped with arrows $p_1:A\times B\to A$, $p_2:A\times B \to B$ and $\pi_1:B\times A\to B$, $\pi_2:B\times A\to A$ respectively by definition of product.
Letting $Y=A\times B$ in the definition of product gives us the arrow $u_1=\langle p_2,p_1\rangle:A\times B\to B\times A$, and similarly $Y=B\times A$, gives us the arrow $u_2=\langle\pi_2,\pi_1\rangle:B\times A\to A\times B$.
By commutativity of the triangles: right including $u_2$; left including $u_2$; left including $u_1$ and right including $u_2$, in the diagram below, we get respectively:
\begin{gather*} p_1\circ u_2=\pi_2,\quad p_2\circ u_2=\pi_1,\quad \pi_1\circ u_1=p_2,\quad \pi_2\circ u_1=p_1 \Rightarrow \\ \pi_2\circ (u_1\circ u_2)=\pi_2,\quad \pi_1\circ (u_1\circ u_2)=\pi_1,\quad p_2\circ (u_2\circ u_1)=p_2,\quad p_1\circ(u_2\circ u_1)=p_1. \end{gather*} Hopefully showing that $u_2\circ u_1=Id_{A\times B}$ and $u_1\circ u_2=Id_{B\times A}$.
I guess I'm using that identities for an object in $\mathbb{C}$ are unique, and thus we would only really need $\pi_1\circ (u_1\circ u_2)=\pi_1$, and $p_1\circ(u_2\circ u_1)=p_1$, to show this; Similar to showing an element is the identity in a group.
Question(s) 1: Is the proof/argument just given, that $A\times B\cong B\times A$, correct; If not, how can I correct it?
Question 2: Would this also show that the products, seen as functors (I guess provided that $\mathbb{C}$ has products) are isomorphic, or is this where naturality comes in, and we actually need to find the natural transformation between the functors?
$\newcommand\C{\mathscr C}$To answer Question 2 if you know adjoint functors: Let consider the following functors: \begin{align} \text{diagonal functor}&&\Delta&:\C\to\C\times\C&X&\mapsto (X,X)\\ \text{switch functor}&&\Sigma&:\C\times\C\to\C\times\C&(X,Y)&\mapsto(Y,X)\\ \text{product functor}&&\Pi&:\C\times\C\to\C&(X,Y)&\mapsto X\times Y \end{align} You are looking for a natural isomorphism $\zeta:\Pi\to\Pi\circ\Sigma$.
First note that $\Sigma$ is an isomorphism of categories since $\Sigma\circ\Sigma$ is the identity functor. Now recall that $\Pi$ is characterized, up to natural isomorphisms, to be the right adjoint of $\Delta$, that's $\Delta\dashv\Pi$. Then $\Sigma\circ\Delta\dashv\Pi\circ\Sigma$, but $\Sigma\circ\Delta=\Delta$, hence $\Delta\dashv\Pi\circ\Sigma$, from which $\Pi\cong\Pi\circ\Sigma$.