Igor wrote:

> Still, what is the shortest and the most correct answer to the puzzle 278, how to formulate it?

'Shortest' and 'most correct' are complementary quantities. Anindya wins the prize for brevity in comment #17:

> \\((f\otimes g)\circ(f'\otimes g') = \otimes(f, g)\circ\otimes(f', g') = \otimes((f, g)\circ(f', g')) = \otimes(ff', gg') = ff'\otimes gg'\\)

and every step here is perfectly correct... but the best possible answer will _explain_ why each step is valid. So, something like this would make me very happy;

**Solution to Puzzle 278.**

Suppose \\(\mathcal{C}\\) is a monoidal category. It is equipped with a functor

\[ \otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}. \]

Since a functor preserves composition, if we take any composable morphisms \\( (f,g) \\) and \\( (f',g') \\) in \\( \mathcal{C} \times \mathcal{C} \\) we must have

\[ \otimes ( (f,g) \circ (f',g')) = \otimes (f,g) \; \circ \; \otimes (f',g') .\]

However, we compose morphisms in \\( \mathcal{C} \times \mathcal{C} \\) componentwise, so

\[ (f,g) \circ (f',g') = (f\circ f', g \circ g'). \]

Thus, the previous equation gives

\[ \otimes ( f \circ f', g \circ g' ) = \otimes (f,g) \; \circ \; \otimes (f',g'). \]

Moreover, we usually write the effect of applying the functor \\( \otimes \\) to a pair \\( (f,g) \\) not as \\( \otimes (f,g) \\) but as \\( f \otimes g\\). If we do this three times above, we get the **interchange law**

\[ (f \circ f') \otimes (g \circ g') = (f \otimes g) \circ (f' \otimes g'). \]

> Still, what is the shortest and the most correct answer to the puzzle 278, how to formulate it?

'Shortest' and 'most correct' are complementary quantities. Anindya wins the prize for brevity in comment #17:

> \\((f\otimes g)\circ(f'\otimes g') = \otimes(f, g)\circ\otimes(f', g') = \otimes((f, g)\circ(f', g')) = \otimes(ff', gg') = ff'\otimes gg'\\)

and every step here is perfectly correct... but the best possible answer will _explain_ why each step is valid. So, something like this would make me very happy;

**Solution to Puzzle 278.**

Suppose \\(\mathcal{C}\\) is a monoidal category. It is equipped with a functor

\[ \otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}. \]

Since a functor preserves composition, if we take any composable morphisms \\( (f,g) \\) and \\( (f',g') \\) in \\( \mathcal{C} \times \mathcal{C} \\) we must have

\[ \otimes ( (f,g) \circ (f',g')) = \otimes (f,g) \; \circ \; \otimes (f',g') .\]

However, we compose morphisms in \\( \mathcal{C} \times \mathcal{C} \\) componentwise, so

\[ (f,g) \circ (f',g') = (f\circ f', g \circ g'). \]

Thus, the previous equation gives

\[ \otimes ( f \circ f', g \circ g' ) = \otimes (f,g) \; \circ \; \otimes (f',g'). \]

Moreover, we usually write the effect of applying the functor \\( \otimes \\) to a pair \\( (f,g) \\) not as \\( \otimes (f,g) \\) but as \\( f \otimes g\\). If we do this three times above, we get the **interchange law**

\[ (f \circ f') \otimes (g \circ g') = (f \otimes g) \circ (f' \otimes g'). \]