Michael wrote in 37:

> The last two questions are still unclear to me...

Remember that in the case of a natural transformation \\(\alpha: F \rightarrow G \\) you have two functors \\(F, G : \mathcal{C} \rightarrow \mathcal {D}\\) and that for objects \\(A\\) and \\(B\\) in \\(\mathcal{C}\\) and morphism \\(f : A \rightarrow B\\) we need to have \\(\alpha_B \circ F(f) = G(f) \circ \alpha_A\\)

In this case we have functors \\(LP, RP: \mathcal{C} \times \mathcal{C} \times \mathcal{C} \rightarrow \mathcal{C} \\) and we need to show that the following sqaures commute:

![](https://docs.google.com/uc?id=1P6DGIbppt9Q3jUJAQhVu9StwvxTNX_uJ)

or with the functors applied:

![](https://docs.google.com/uc?id=1LzqagB1kwvwwKQuUO_Z1r_tbYqgqI_C1)

Also, remember how the product functor acts on morphisms:

![](https://docs.google.com/uc?id=1wq7b12xeRVIGfpY6SqOFiqf1qzt6XGJd)

With these hints and your projection diagram, you should be able to solve the problem.

> The last two questions are still unclear to me...

Remember that in the case of a natural transformation \\(\alpha: F \rightarrow G \\) you have two functors \\(F, G : \mathcal{C} \rightarrow \mathcal {D}\\) and that for objects \\(A\\) and \\(B\\) in \\(\mathcal{C}\\) and morphism \\(f : A \rightarrow B\\) we need to have \\(\alpha_B \circ F(f) = G(f) \circ \alpha_A\\)

In this case we have functors \\(LP, RP: \mathcal{C} \times \mathcal{C} \times \mathcal{C} \rightarrow \mathcal{C} \\) and we need to show that the following sqaures commute:

![](https://docs.google.com/uc?id=1P6DGIbppt9Q3jUJAQhVu9StwvxTNX_uJ)

or with the functors applied:

![](https://docs.google.com/uc?id=1LzqagB1kwvwwKQuUO_Z1r_tbYqgqI_C1)

Also, remember how the product functor acts on morphisms:

![](https://docs.google.com/uc?id=1wq7b12xeRVIGfpY6SqOFiqf1qzt6XGJd)

With these hints and your projection diagram, you should be able to solve the problem.