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?