Michael Hong, as always an excellent drawing, thank you!

Concerning

\\(\otimes((f \circ f'),(g \circ g')) = \otimes(f,g) \circ \otimes(f',g')\\)

1. We have a product \\(C \times C\\), which has as objects ordered pairs \\((x, y)\\), and morpisms between such pairs, also ordered pairs like \\((f, g)\\), so \\(C \times C((x, y), (x', y')) = C(x, x') \times C(y, y'))\\). This is all we know from definition.

2. Let's say the functor \\(\otimes\\) maps an ordered pair \\((f, g) \\) to some morphism \\(a: C \to C\\), and \\((f', g')\\) to \\(b: C \to C\\), like this: \\(\otimes(f, g) = a\\) and \\(\otimes(f', g') = b\\), and of course preserves composition \\(\otimes((f' g') \circ (f, g)) = \otimes((f',g')) \circ \otimes((f,g))\\).

What I need to figure out what is \\((f' ,g') \circ (f, g)\\). We have a lot of morphisms to choose from - \\(C \times C((x, y), (x'', y'')) = C(x, x'') \times C(y, y''))\\), and among them, by definition of category, we have a morphism \\((f' \circ f, g' \circ g)\\). So we picked it up as the only candidate for \\((f', g') \circ (f, g)\\), which makes perfect sense, but still I'm not sure how to show this.

Concerning

\\(\otimes((f \circ f'),(g \circ g')) = \otimes(f,g) \circ \otimes(f',g')\\)

1. We have a product \\(C \times C\\), which has as objects ordered pairs \\((x, y)\\), and morpisms between such pairs, also ordered pairs like \\((f, g)\\), so \\(C \times C((x, y), (x', y')) = C(x, x') \times C(y, y'))\\). This is all we know from definition.

2. Let's say the functor \\(\otimes\\) maps an ordered pair \\((f, g) \\) to some morphism \\(a: C \to C\\), and \\((f', g')\\) to \\(b: C \to C\\), like this: \\(\otimes(f, g) = a\\) and \\(\otimes(f', g') = b\\), and of course preserves composition \\(\otimes((f' g') \circ (f, g)) = \otimes((f',g')) \circ \otimes((f,g))\\).

What I need to figure out what is \\((f' ,g') \circ (f, g)\\). We have a lot of morphisms to choose from - \\(C \times C((x, y), (x'', y'')) = C(x, x'') \times C(y, y''))\\), and among them, by definition of category, we have a morphism \\((f' \circ f, g' \circ g)\\). So we picked it up as the only candidate for \\((f', g') \circ (f, g)\\), which makes perfect sense, but still I'm not sure how to show this.