I am trying to understand the steps in this proof of the fact that products are unique.
$$\begin{array}{center}
\; A \times B \\
\; p_1 \swarrow \, \, \, \, \, \, \downarrow f \, \, \, \, \, \, \searrow p_2 \\
\; A \longleftarrow \; A \times' B \longrightarrow B \\
\end{array}$$
By the universal property of products, there is a unique morphism $f$ such that the diagram above commutes.
Question 1: Can I use the universal property again in the diagram above to get a unique morphism $g: A \times'B \to A \times B$ such that the diagram commutes and hence they are isomorphic? I believe not, because the universal property only makes this diagram commutes:
$$\begin{array}{center}
\; A \times' B \\
\; p_1 \swarrow \, \, \, \, \, \, \downarrow g \, \, \, \, \, \, \searrow p_2 \\
\; A \longleftarrow \; A \times B \longrightarrow B \\
\end{array}$$
and so my argument does not work.
The proof in the article then proceeds to "glue" the two together so that $g$ makes the entire "glued together" diagram commutes. That is, $p_1 \circ (g \circ f) = p_1$.
Question 2: Why does the universal property guarantee a $g$ such that $p_1 \circ (g \circ f) = p_1$? When can we join commutative diagrams together and then apply the universal property?
Best Answer
Basically, you are trying to prove that for any two objects $A$ and $B$, if the product $A\times B$ exists then it is unique up to unique isomorphism. Basically, there may be some other $A\times'B$, but then there exists a unique isomorphism from $A\times B$ to $A\times' B$ which commutes with the projections. Remember the important definition, that an isomorphism $f: A \to B$ is an arrow such that there exists an arrow $g$ where the two identities $g\circ f = 1_A$ and $f\circ g = 1_B$.
If $A\times B$ and $A\times' B$ both satisfy the definition of the product of $A$ and $B$ then we have the commutative diagrams:
$$\begin{array}{center} \; A \times B \\ \; p_1 \swarrow \, \, \, \, \, \, \downarrow f \, \, \, \, \, \, \searrow p_2 \\ \; A \longleftarrow \; A \times' B \longrightarrow B \\ \;p'_1~\;\;\;\;\;\;\;\;\;\;~\;p'_2\\ \end{array}$$
and
$$\begin{array}{center} \; A \times' B \\ \; p'_1 \swarrow \, \, \, \, \, \, \downarrow g \, \, \, \, \, \, \searrow p'_2 \\ \; A \longleftarrow \; A \times B \longrightarrow B \\ \;p_1~\;\;\;\;\;\;\;\;\;\;~\;p_2\\ \end{array}$$
Where the arrows $f$ and $g$ must be unique by the definition of the product. Composing them also gives the unique morphism $g\circ f: A\times B \to A\times B$ which satisfies the universal property of the product again. But we also have that $1_{A\times B}$ is another arrow making the required diagram commute, so they must be identical, i.e. $g\circ f = 1_{A\times B}$. The same argument with the two products reversed shows that $f\circ g = 1_{A\times'B}$. Therefore there is an isomorphism between the two products and it is unique because the arrows $f$ and $g$ are. This argument is summarized by that "glued together" diagram on the website.