And if I write it like this, I want this i to be a functor.

The reason why we said that there is a real parameter category behind it and we don't just indicate it,

is that I want this to be a functor.

So this initial algebra is functionally dependent on the parameter.

So that means we gave a re-indication f from a to b in a.

I said f, I mean what I write, namely p, just so that these parametermorphisms are somehow called parameters and not morphisms in the basic category.

So now I want to define Ip as a c-morphism from Ia to Ib.

So the chance I have to do this is of course to use the initiality of Ia.

So I have this initial algebra structure Iota a, we wanted to write it.

So f from Ia, a to Ia.

Because again, this is an f, a algebra, that means I put my underlying amount here again.

We'll do it again, the same is true for Ib, an initial f, b algebra, that means we have this Iota b, which goes from f from Ib, b to Ib.

So, it doesn't quite fit yet, so to get this Ip as a morphism from an initial f, a algebra, we would have to make this thing an f, a algebra.

How do we do that? Well, f, a algebra means we have f applied to our underlying object, i.e. to Ib, and at the second place the a.

So, we have to get here somehow. Now I didn't paint it nicely, too bad. Not nice in the sense that the name of this morphism doesn't fit well here anymore, but we all see which one it is.

I mean, you don't have to think of one that works.

The homfunktor is difficult, it is a two-digit homfunktor, but it is not a homfunktor in general. A homfunktor would be a counter-variant in the first argument, but this one is a co-variant. So, the homfunktor is not even a candidate.

Ooops...

How do we get from here to there?

As I said, there is something.

The IP doesn't have to depend on P.

So, you have to get a little bit of it.

The B-Functor in two arguments is basically an instance of a normal Functor.

A B-Functor is a Functor on the product category.

But you have to be very clear about how it works with the application.

You have to get used to thinking in a multistage manner.

I can always hold one part and use the other.

Of course, I can also use both sides of morphism.

In general, this is the application of F on a single morphism in the product category.

The product category consists of pairs.

In the product category C-A, I have an object.

A pair of objects, one from C, one from A.

And the morphisms are also pairs.

Here I have for example ID on IB, P.

And I remind you that we can use the name of the object as a abbreviation for the identity of the object.

This is the same as writing F of ID, P.

This is the name of the object as an ID on IB.

In what sense is the name of the object as an ID?

The object is a binary object.

The product of vectors is known even if you do not know what a vector space is.

The scalar product is bilinear.

If the product of vectors is a scalar product of V plus W, then it is a scalar product of V, U plus a scalar product of W, U.

This is the scalar product of the product.

It is bilinear.

It is not linear.

It is not true.

We have very different situations here.

This is what has a significant influence on the structure of categories.

If you have heard of the tensor product of vector spaces, you might be the only person in the room who thinks so.

Tensors are important.