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 \$$