Great answer to Puzzle 278, Michael!

Igor wrote:

> What I need to figure out what is \\((f' ,g') \circ (f, g)\\).

Okay, you're trying to figure out how to compose two morphisms in \\(\mathcal{C} \times \mathcal{C}\\). For this you can look at [Lecture 52](https://forum.azimuthproject.org/discussion/2273/lecture-52-the-hom-functor/p1), where I explained how to compose morphisms in a product of categories:

> **Theorem.** For any categories \\(\mathcal{X}\\) and \\(\mathcal{Y}\\), there is a category \\(\mathcal{X} \times \mathcal{Y}\\) for which:

> * An object is a pair \\( (x,y) \in \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{Y}) \\).

> * A morphism from \\( (x,y) \\) to \\( (x',y') \\) is a pair of morphisms \\( f: x \to y'\\) and \\(g: y \to y'\\). We write this as \\( (f,g) : (x,y) \to (x',y') \\).

> * We compose morphisms as follows:

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

> * Identity morphisms are defined as follows:

> \[ 1_{(x,y)} = (1_x, 1_y) .\]

> **Proof.** Just check associativity and the right/left unit laws. \\( \qquad \blacksquare \\)

Igor wrote:

> What I need to figure out what is \\((f' ,g') \circ (f, g)\\).

Okay, you're trying to figure out how to compose two morphisms in \\(\mathcal{C} \times \mathcal{C}\\). For this you can look at [Lecture 52](https://forum.azimuthproject.org/discussion/2273/lecture-52-the-hom-functor/p1), where I explained how to compose morphisms in a product of categories:

> **Theorem.** For any categories \\(\mathcal{X}\\) and \\(\mathcal{Y}\\), there is a category \\(\mathcal{X} \times \mathcal{Y}\\) for which:

> * An object is a pair \\( (x,y) \in \mathrm{Ob}(\mathcal{X}) \times \mathrm{Ob}(\mathcal{Y}) \\).

> * A morphism from \\( (x,y) \\) to \\( (x',y') \\) is a pair of morphisms \\( f: x \to y'\\) and \\(g: y \to y'\\). We write this as \\( (f,g) : (x,y) \to (x',y') \\).

> * We compose morphisms as follows:

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

> * Identity morphisms are defined as follows:

> \[ 1_{(x,y)} = (1_x, 1_y) .\]

> **Proof.** Just check associativity and the right/left unit laws. \\( \qquad \blacksquare \\)